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TEMPORARY STRUCTURE DESIGN

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TEMPORARY STRUCTUREDESIGN

CHRIS SOUDER

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Cover image: © Chris Souder

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I would like to dedicate this book to Stuart (Bart) Bartholomew.

I only knew Bart for 14 years, but in this short time he had more

influence on me than most. Not only did Bart design the temporary

structure class that this book was designed for, he was instrumental

in my decision to change to the teaching profession. Countless breakfast

and lunch meetings had me listening in amazement to the years

of construction, teaching, and consulting experiences. Bart, you are a

model of integrity, honesty, and ethical behavior.

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CONTENTS

ABOUT THE AUTHOR xiii

PREFACE xv

ACKNOWLEDGMENTS xvii

1 Statics Review 1

1.1. Statics Review / 1

1.2. Units of Measure / 1

1.2.1. Common Units of Measure / 2

1.3. Statics / 3

1.3.1. Centroids/Center of Gravity / 4

1.3.2. Properties of Sections / 7

2 Strength of Materials Review 18

2.1. Stress / 18

2.1.1. Normal Stress / 18

2.1.2. Bending Stress / 19

2.1.3. Shear Stress / 19

2.1.4. Horizontal Shear Stress / 20

2.1.5. Modulus of Elasticity / 22

vii

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viii CONTENTS

2.2. Bending Moments / 22

2.2.1. Maximum Bending Moments / 22

2.2.2. Maximum Shear / 23

2.2.3. Law of Superposition / 23

2.3. Materials / 24

2.3.1. Factors of Safety / 24

2.3.2. Grades of Steel / 24

2.3.3. Compact Beam / 25

2.3.4. Wood / 26

2.4. Deflection / 27

2.5. Shear and Moment Diagrams / 28

2.6. Beam Design / 34

2.6.1. Combined Stress / 41

3 Types of Loads on Temporary Structures 45

3.1. Supports and Connections on Temporary Structures / 45

3.1.1. Forces and Loads on Temporary Structures / 47

3.1.2. Materials—How Different Materials Create Different

Forces / 48

4 Scaffolding Design 59

4.1. Regulatory / 59

4.2. Types of Scaffolding / 59

4.3. Loading on Scaffolding / 61

4.4. Scaffolding Factors of Safety / 62

4.5. Scaffold Components / 62

4.5.1. Planking / 62

4.5.2. Bearers (Lateral Supports) / 62

4.5.3. Runners / 62

4.5.4. Posts / 63

4.5.5. OSHA / 63

4.6. Scaffold Design / 63

4.6.1. Securing Scaffolding to the Structure / 69

4.6.2. Hanging Scaffold / 69

5 Soil Properties and Soil Loading 75

5.1. Soil Properties / 75

5.1.1. Standard Penetration Test and Log of Test Borings / 77

5.1.2. Unit Weights above and below the Water Table / 78

5.1.3. Testing / 81

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CONTENTS ix

5.2. Soil Loading / 81

5.2.1. Soil Mechanics / 81

5.2.2. Active Soil Pressure and Coefficient / 82

5.2.3. Soil Pressure Theories / 83

5.2.4. Soil Pressure Examples Using Rankine Theory / 85

5.2.5. Soil Pressures Using State and Federal Department

Standards / 91

6 Soldier Beam, Lagging, and Tiebacks 104

6.1. System Description and Units of Measure / 104

6.1.1. Beams/Piles / 104

6.1.2. Lagging / 105

6.1.3. Tiebacks / 105

6.2. Materials / 105

6.2.1. Steel AISC / 105

6.2.2. Wood Species—National Design Specifications (NDS)

for Wood Construction / 106

6.2.3. Lagging / 108

6.2.4. Soldier Beam Design / 112

6.2.5. Tiebacks and Soil Nails / 121

7 Sheet Piling and Strutting 130

7.1. Sheet Piling Basics / 130

7.1.1. Materials / 130

7.1.2. System Description and Unit of Measure / 130

7.1.3. Driving Equipment / 133

8 Pressure and Forces on Formwork and Falsework 155

8.1. Properties of Materials / 155

8.1.1. Unit Weights / 155

8.1.2. Forces from Concrete Placement / 157

9 Concrete Formwork Design 178

9.1. General Requirements / 178

9.1.1. Concrete Specifications / 178

9.1.2. Types and Costs of Forms in Construction / 179

9.2. Formwork Design / 180

9.2.1. Bending, Shear, and Deflection / 180

9.2.2. Form Design Examples Using All-Wood Materials

with Snap Ties or Coil Ties / 191

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x CONTENTS

9.2.3. Formwork Charts / 199

9.2.4. Estimating Concrete Formwork / 219

9.3. Conclusion / 228

10 Falsework Design 229

10.1. Falsework Risks / 229

10.1.1. Falsework Accidents / 230

10.1.2. Falsework Review Process / 233

10.1.3. Falsework Design Criteria / 235

10.1.4. Load Paths for Falsework Design / 236

10.1.5. Falsework Design Using Formwork Charts / 242

10.1.6. Bridge Project / 262

11 Bracing and Guying 267

11.1. Rebar Bracing and Guying / 268

11.2. Form Bracing with Steel Pipe and Concrete Deadmen / 269

11.2.1. Life Application of Friction Forces / 278

11.3. Rebar Guying on Highway Projects / 279

11.4. Alternate Anchor Method / 289

12 Trestles and Equipment Bridges 300

12.1. Basic Composition of a Standard Trestle / 300

12.1.1. Foundation—Pipe, H Pile, and Wide-Flange and Composite

Piles / 301

12.1.2. Cap Beams—Wide-Flange Beams with Stiffeners / 301

12.1.3. Stringers/Girders—Wide-Flange Beams Braced Together / 303

12.1.4. Lateral Bracing / 303

12.1.5. Decking—Timber or Precast Concrete Panels / 306

12.1.6. Environmental Concerns / 308

12.1.7. Stringer Design / 325

12.1.8. Star Pile Design and Properties / 340

12.2. Other Projects Utilizing Methods of Access / 341

12.3. Conclusion / 343

13 Support of Existing Structures 344

13.1. Basic Building Materials / 345

13.1.1. Example 13.1 Pipe Unit Weight / 346

13.1.2. Example 13.2 Existing Water Treatment Plant / 347

13.1.3. Example 13.3 Temporary Pipe Supports / 354

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CONTENTS xi

Appendixes 369

Appendix 1: Steel Beams (AISC) 371

Appendix 2: Steel Pipe 391

Appendix 3: H Pile (AISC) 393

Appendix 4: Allowable Buckling Stress 395

Appendix 5: Sheet Pile (Skyline) 397

Appendix 6: Wood Properties 401

Appendix 7: Formwork Charts (Williams) 404

Appendix 8: Form Hardware Values (Williams) 412

Appendix 9: Aluminum Beams (Aluma) 422

INDEX 425

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ABOUT THE AUTHOR

Chris Souder graduated with an undergraduate degree in construction management

in 1988 before going to work for Kiewit Pacific Co. in northern California. Chris had

a successful 16-year career with Kiewit and was involved with many projects in the

heavy civil arena. Chris held positions from field engineer to project manager to lead

estimator. Some of the projects Chris was involved with were the Woodland WWTP

expansion in Woodland, California, Highway 85 Bridge construction for CalTrans in

San Jose, California, WWTP Expansion and new facilities for the City of Roseville

at its Booth Rd. and Pleasant Grove Plants, Highway 101 Retrofit work for CalTrans

in San Francisco, California, new Highway 880 construction of bridge structures for

CalTrans in Oakland, California, following the 1989 Loma Prieta earthquake, Water

storage facilities for the City of Sacramento, new bridge and 2 miles of road construc-

tion including a pump station in Oroville, California, an expansion of the Sacramento

River WTP facility for the City of Sacramento, and various estimating assignments

for both heavy highway and water treatment facilities throughout northern California.

These projects as a whole had total revenues in excess of $420 million.

Chris then pursued an Interdisciplinary Master’s degree in construction planning

at California State University, Chico, while teaching full time in the construction

management program. Today, Chris teaches temporary structures and scheduling and

project controls to fourth-year students at Chico State while maintaining a continuous

portfolio of consulting projects and industry trainings ranging from cost estimating,

temporary structures design, and scheduling services. While teaching, Chris received

the terminal degree in construction management by completing his M.S. in construc-

tion planning at Chico State. This education, combined with 16 years of heavy civil

industry experience makes Chris a most effective type of professor in the construction

management discipline.

xiii

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PREFACE

Temporary structure design is not taken lightly by the owner, engineer, or contractor.

It has and should always be a practice that is performed by a licensed engineer in its

specific discipline. However, the construction manager should be versed in the design

procedures to a point where he can request a particular design or review a concept or

submittal with the ability to understand the basic components of the design.

In 1989, the fourth edition of Simplified Mechanics and Strength of Materials waswritten. This book is an example of the present book’s goal. I was inspired by the

simplicity that Parker and Ambrose displayed in their text. I truly believe that this

subject can be well understood by the construction manager without the ultimate

goal of becoming a licensed engineer. However, if that is the goal of the student, this

text will prepare you to take the next step in engineering pursuing your goal to be

licensed.

There is a need for this topic in a construction management (CM) degree, both

undergraduate and graduate, civil engineering (CE), both under graduate and grad-

uate, or in industry that is simplified enough that the student, intern, or engineer

can simply follow the major concepts without sacrificing key engineering principles.

Different universities approach the temporary structures topic in several ways. Some,

like Chico State, make it the culminating experience following statics and mechanics.

This text will compliment a similar program. Others teach a “structure” class that

gives the students a basic understanding of how structures are designed. The latter

focuses more on permanent design. Many civil engineering students graduate and

go on to work for state agencies or heavy civil contractors. Both of these careers

rely heavily on the design of temporary structures. With the state agency, one will

be reviewing and inspecting temporary structures. With the contractor, one will be

involved with helping design and building temporary structures. These two paths are

very rewarding for a CM or CE undergraduate or graduate student.

xv

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xvi PREFACE

I also wanted students of temporary structures to be able to comprehend the more

complicated analysis that come with more difficult loading conditions without the

need for a complete understanding or need for indeterminate structure analysis. I want

the student to be aware of the available software on the market today that can simplify

even the most complicated loading condition.

I also thought it was important that the student or engineer of this subject be able

to understand and perform simple cost estimates of the designs that are explained in

each chapter. Most chapters have brief explanations of cost analyses so the educated

decisions can be made during the design phase.

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ACKNOWLEDGMENTS

To Jessica, Jason and Devon, thank you for putting up with me through this life of

construction. To my brothers Greg and Mike, Greg for encouraging me to go into a

construction management program andMike for his constant support. To my parents,

Dick and Ines Souder for your constant support, even if you really don’t know what

I do for a living.

To Robert Towne and Ruth Younger for your hours of assistance to the manuscript.

To Valentina Pozin and Don Hamann for your sound engineering advise and

“hand holding.”

To the following industry folks and colleagues:

Jim Dick

Steve Floyd

Howard Mattfield

Jim Cole

Dan Collins

Jon Re

Dave Mitchell

Clint Cole

Ken Riley

Dave Jack

Dan Griffin

Shawn Drobny

Brad Kaufman

Bill Cooke

xvii

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xviii ACKNOWLEDGMENTS

Matt Halleen

Dan Munson

Dave Hazen

Rovane Younger

Bruce Yoakum

Lee Cushman

for your contributions, advice, and case study assistance.

To the following companies:

Kiewit

Traylor Bros.

Golden State Bridge

Cushman Contractors Inc.

Flatiron Corporation

Pankow Builders, LTD

for your case study contributions.

I thank you all.

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TEMPORARY STRUCTURE DESIGN

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CHAPTER 1

STATICS REVIEW

1.1 STATICS REVIEW

In construction management and civil engineering programs, students are required

to take statics and strength of material classes in preparation for their successor. The

successor might be a generic “structures” course, a temporary structure course, or

maybe no successor course at all. Whichever direction the curriculum goes, the basics

of statics and strength of materials is the common denominator.

This book has been written under the assumption that the student has a background

in statics and strength of materials and these skills only need to be refined. Temporary

structures utilizes many of the less complicated aspects of statics and strength of

materials, so even if the student did not master the two prerequisites, he should still be

successful in the subject matter of this book. In addition, temporary structure design is

a very practical subject, and the student should be energized to see that the challenges

that this book covers are real construction situations that the student will experience

for his or her entire career.

1.2 UNITS OF MEASURE

At the time of this writing, local and state projects in the United States continue to

use the English “Imperial” unit system (feet, pounds, etc.). While most of Europe

and the rest of the world use the metric system, the United States has resisted this

movement. Even the California Department of Transportation, which had converted

current and future projects to the Imperial system of measures late in the 20th century,

has gone back to using the Imperial system in the early 21st century. Since England

1

2 STATICS REVIEW

TABLE 1.1 Units of Measure

Unit Name Unit of Measure

LengthFoot ft (′)

Inches in (′′)

AreaSquare feet SF, ft2

Square inches in2

VolumeCubic feet CF, ft3

Cubic inches in3

Force and PressurePound lb, #

Kip k (1000 lb)

Pounds per ft lb/ft

Kips per ft k/ft

Pounds per SF lb/SF, psf

Pounds per linear foot lb/ft

Kips per linear foot kpf

Kips per SF k/SF, ksf

Pounds per CF lb/CF, pcf

MomentFoot-pounds ft-lb

Inch-pounds in-lb

Foot-kips ft-k

Inch-kips in-k

StressPounds per ft2 psf, lb/ft2

Pounds per in2 psi, lb/in2

Kips per ft2 ksf, k/ft2

Kips per in2 ksi, k/in2

TemperatureDegree Fahrenheit ∘F

has also gone to the metric system, their “English” Imperial system is now referred to

as the U.S. units. Because this text has been written for students in the United States,

examples will be given in U.S. units only. Table 1.1 shows most of the common units

of measure used in this book.

1.2.1 Common Units of Measure

With any engineering subject, the use of variables to represent different engineering

values is standard. These symbols derive either from the Greek alphabet or English

letters. Regardless, a great number of symbols are necessary to represent the vari-

ous engineering concepts. Table 1.2 shows the notations and symbols most used in

this book.

1.3 STATICS 3

TABLE 1.2 Notation and Symbols

Subject Symbol (Variable) Description

Properties S Section modulus

I Moment of inertia

E Modulus of elasticity

A Cross-sectional area

r Radius of gyration

R Radius of a circle

e Eccentricity

a Moment arm distance

b Beam width

c Distance from centroid to top or bottom edge

d Depth of beam

D Diameter of a circle

g Acceleration of gravity

h Height or depth

K Distance from top of beam to tangent of web

e Effective length of a column or strut, distance

from top of beam to web tangent

Stress fb Bending stress

fv Shear stress

f ′v Twice shear stress used for short-term shear

loading

fc∥ Normal compression stress parallel to the

grain of the wood

fc Normal compression stress perpendicular to

the grain of the wood

Soil Mechanics c Cohesion (psf)

𝜙 Angle of internal friction (degrees)

𝛽 Passive slip plane angle (degrees)

𝛼 Active slip plane angle (degrees)

𝜇 Coefficient of friction

𝛾 Unit weight (pcf)

𝜋 𝜋 = 3.1416

ka Active soil coefficient

kp Passive soil coefficient

T Temperature

1.3 STATICS

Statics is the study of an object that is not moving, hence static or equilibrium. A force

is a motion or change of motion in a body. A common force that is produced on Earth

naturally is gravity. Gravity is the tendency of the weight of a body to be attracted to

the center of the Earth. The mass of some unit weight is placed in motion by gravity

or other means. The force of the mass originates at the center of gravity of the body

in question. Thus, there is direction and a known weight. Another way to describe a

force is something that has magnitude and direction.

4 STATICS REVIEW

FIGURE 1.1 Force not centered on beam.

Active forces are those forces created by the magnitudes and directions (P) such asthe resultant force of a load of concrete. Reactions (R) are also forces, but they are theby-product of the sum of the resultant forces, such as two beams supporting another

beam on each of its ends. For instance, Figure 1.1 shows an active P (10 k) force

acting downward on the beam, 4 ft from the left side. The reactions are the forces at

the supports pushing back (upward). The reactions between the left and right side are

not the same because the active force is not centered on the beam.

Forces can be in line with the longitudinal axis of a member or act eccentrically.

Axial, compression, and tension forces typically act in line with the linear axis of the

member. Eccentric loading introduces forces coming from different directions from

the normal forces. The forces are not always vertical or horizontal but sometimes

act in directions that are not normal to either axis. Standard structural members are

designed to best transfer forces along their linear axes; however, we are not always

able to comply with this optimum design.

For the purpose of temporary structures, forces are either uniform (linear) or con-

centrated. Concentrated loads are also known as point loads and are represented by

force arrows and force value in pounds or kips as shown above. Linear loads come

in different forms. They can be triangular, trapezoidal, or rectangular depending on

the material creating the forces. These forces are represented by a sequence of arrows

and a value in pounds per linear foot (lb/ft) or kips per linear foot (kips/ft) as shown in

Figure 1.2. Different types of forces created in temporary structures will be discussed

in Chapter 3.

1.3.1 Centroids/Center of Gravity

When moment of inertia is introduced to the statics student, it is typically discussed

after center of gravity is understood. The student should spend time understanding

where and why each object has a center of gravity. In this book, the center of gravity

(COG) and “centroid” or “centroidal axis” is used synonymously.

The center of gravity of a square, circle, or rectangle is in the center of each dimen-

sion of a material of uniform density. The vertical is referred to as the y axis and

the horizontal is referred to the x axis. The center of gravity of circles (pipes) and

rectangles is illustrated in Figure 1.3.

FIGURE 1.2 Uniform load on beam.

1.3 STATICS 5

FIGURE 1.3 Center of gravity of pipes and rectangles.

FIGURE 1.4 Center of gravity of a composite shape.

Shapes that are not symmetrical have centers of gravity that are also not symmet-rical. For instance, if the shape in Figure 1.4 is considered, where would the center

of gravity be in the x and in the y axis?

Example 1.1 If A has a base of 4′′ and a height of 6′′, and B has a base of 4′′ and aheight of 3′′ and the top of A and B are flush, determine the composite’s COG in the

x and y directions.

Step 1: Calculate the area ofA andB and determine each shape’s individual center

of gravity:

A = 4′′ × 6′′ = 24 in2

A’s COG is half the base and half the height, 2′′ and 3′′:

B = 4′′ × 3′′ = 12 in2

B’s COG is half the base and half the height, 2′′ and 1.5′′. The sum of the

two areas is 24 + 12 = 36 in2.

Step 2: Determine the distance of the x axis to the bottom of the composite by

adding the moments of each about the bottom plane. Figure 1.5 illustrates

this dimension.

FIGURE 1.5 Center of gravity determining y′.

6 STATICS REVIEW

A’s COG is 1∕2 (6′′) = 3′′ from the bottom.

B’s COG is 3′′ + 1∕2 (3′′) = 4.5′′ from the bottom.

Now considering their areas, total the areas about the bottom and make

them equal to the combined area and its unknown distance to the new

centroid (y′). This will be how far upward the horizontal centroid is:

[(24 in2)(3′′)] +[(12 in2

)(4.5′′)

]= [(36 in2)(y′)]

Solve for y′:

y′ = (24 in2)(3′′) + (12 in2)(4.5′′)36 in2

y′ = 3.5′′

Step 3: Determine the distance of y′ from the bottom of the composite by sum-

ming the moments of each about the left side. Figure 1.6 illustrates this

dimension.

A’s COG is 1∕2 (4′′) = 2′′ from the left.

B’s COG is 4′′ + 1∕2 (4′′) = 6′′ from the left.

Once again considering their areas, sum the areas about the left side and

make them equal to the combined area and its unknown distance to the

new centroid (x′). This will be how far to the right the vertical centroid is:

[(24 in2)(2′′)] + [(12 in2)(6′′)] = [(36 in2)(x′)]

Solve for x′:

x′ = (24 in2)(2′′) + (12 in2)(6′′)36 in2

x′ = 3.33′′

FIGURE 1.6 Center of gravity determining X.

Figure 1.7 shows the solution graphically.

FIGURE 1.7 Center of gravity determining solution.

1.3 STATICS 7

Figure 1.8 shows some sample calculations for some common shapes.

×

π

FIGURE 1.8 Centroids of area of common shapes.

1.3.2 Properties of Sections

Cross-Sectional Area The cross-sectional area of a beam consists of the depthtimes the width or the combination of these if there is an irregular shape. For a rect-angular or square section, it is defined as

A = b × d

where b = beam width

d = beam depth

Centroidal Axis The centroid of a beam section is important as one begins to studymoment of inertia. The centroid is the center of gravity of the beam and can be mea-sured about the x or the y axis. The distance from the centroid to the outermost surfaceof a beam is important because this is the location where the highest stress occurs.When a beam is placed in a bending condition, its stress is measured by the beamssection modulus value. The section modulus is a derivative of the beams moment ofinertia and the distance from the centroid to the outermost fiber (surface).

Moment of Inertia Moment of inertia is introduced in statics following theunderstanding of centroids and center of gravity. These basic concepts are key tounderstanding how beam sections are influenced by how far their sections extendbeyond the centroidal axis. Moment of inertia of beams increases as the beamscross-sectional properties are farther from the centroidal axis. Moment of inertia isused to determine deflection in beams and is paramount to the value of section modu-lus, a very important component of a beam’s bending resistance (to be covered in thenext section).

8 STATICS REVIEW

The moment of inertia for a rectangle or square section can be solved by the fol-lowing simple formula that is also referred to as the first moment:

I = bd3

12

where b = beam width (in)

d = beam depth (in)

For composite sections, the moment of inertia is determined by the sum of the firstand secondmoments. The first moment is themoment of inertia of the individual com-ponents of the composite (web and flanges). A composite section consists of differentshapes (usually rectangles) that are put together to create one common unit madeof the same material in most cases. The top and bottom horizontal components arereferred to as flanges and the vertical components are referred to as webs or webbing.The second moment takes into account the distance of the composite componentsto the centroid. This is referred to by engineers as the parallel axis theorem. Largemoments of inertia are developed when the individual components become fartherfrom the centroid. Therefore, a tall beam would have a higher moment of inertia asthe top and bottom flange get farther from the centroid. The moment of inertia of acomposite beam can be solved with the following formula referred to by engineersas the parallel axis theorem:

I = ΣI + ΣAd2

where I = first moment (in4)

A = area of an individual component (flange or web)

d = distance from the centroid to the center of the individual component

Example 1.2 Determine the moment of inertia about the horizontal axis for the

composite beam section shown in Figure 1.9.

Beam properties:

The thickness of the flange and the web are both 2′′.

The flange is 6′′ wide (A).

The web is 8′′ tall, not including the flange thickness (B).

FIGURE 1.9 Section drawing of composite section.

Step 1: Determine the centroid (COG) about the horizontal axis.

To do this, total the moments of each section A and B about the bottom

of the composite. Refer to Figure 1.10.

1.3 STATICS 9

FIGURE 1.10 Section drawing of composite section.

Area of A = 2 × 6 = 12 in2

Area of B = 2 × 8 = 16 in2

Distance from bottom to center of A:

YA = 8 + 1 = 9′′

Distance from bottom to center of B:

YB = 8

2= 4′′

(28)(y′) = (12)(9) + (16)(4); y′ = 6.14′′

Step 2: Determine the first and second moments using Table 1.3. Calculate d,which is the distance from the section to the centroid. Table 1.3 sum-

marizes the different values that make up the moment of inertia. It is

recommended that one always use a similar table to organize the infor-

mation to avoid mistakes.∑I +

∑Ad 2 = 260.75 in4

The first and second moments add up to 260.75 in4.

TABLE 1.3 Moment of Inertia Summary

Section Area d (in) Ad2 (in4) I (in4) I + Ad2 (in

4)

A 12 9 − 6.14 = 2.86 98.15 4 102.15

B 16 6.14 − 4 = 2.14 73.27 85.33 158.60

Section Modulus Section modulus is a derivative of moment of inertia. Itincreases or decreases based on the distance from its centroid to its outermost fiber,either top or bottom of the beam section. Section modulus is used when determiningbending stress of a beam. Section modulus can be derived by the following formulaor obtained from a beam section (see Appendix 1). This appendix comes from valuescalculated in the Manual of Steel Construction (MSC). The authors of the MSChave calculated all the beam properties for all available beams manufactured in theUnited States and some other countries. For wood sections, the National DesignSpecification for Wood Construction (NDS) provides values for the most commonwood sections provided. Appendix 6 illustrates these values, which are rectangular

10 STATICS REVIEW

FIGURE 1.11 Distance to extreme fibers.

and square sections. If the moment of inertia and the centroid are known, then thesection modulus can be derived from both. Figure 1.11 illustrates the distance c fromeither top or bottom of the composite.

S = Ic

where I is the moment of inertia and c is the distance from the centroid to the outer-most fiber of the beam. If the composite beam in Example 1.1 was used as a simplysupported beam, the top would be in compression and the bottomwould be in tension.The beam height is 10 in, the center of gravity is 6.14′′ from the bottom. Therefore,the center of gravity is (10 − 6.14) = 3.87′′ from the top.

Based on the section modulus formula above:

Sbottom = 260.75 in3

6.14′′= 42.50 in3

Stop =260.75 in3

10′′ − 6.14′′= 67.55 in3

For a rectangular or square section, since the centroid is in the center of the section,the section modulus is as follows:

S = bd2

6

where b = beam width

d = beam depth

Example 1.3 Determine the section modulus of the section shown where the base

dimension is 3′′ and the depth of the section is 9′′. Refer to Figure 1.12 for details.

FIGURE 1.12 Vertical rectangular section.

1.3 STATICS 11

Since S = bd2∕6, therefore, S = (3)(9)2∕6 = 40.5 in2.

And since I = bd3∕12, therefore, I = (3)(9)3∕12 = 182.3 in4.

If this section is turned 90∘ and its base becomes 9′′ and depth becomes 3′′, thiswould dramatically change the beam’s section modulus (and moment of inertia).

Let’s recalculate the section modulus of the beam in the weak direction. Refer to

Figure 1.13 for details.

FIGURE 1.13 Horizontal rectangular section.

Since S = bd2∕6, therefore, S = (9)(3)2∕6 = 13.5 in3.

And since I = bd3∕12, therefore, I = (9)(3)3∕12 = 20.25 in4.

The bending resistance (S) between the strong axis and the weak axis is three times

higher. The deflection and buckling resistance (I) between the strong and the weak

axes is nine times higher.

Modulus of Elasticity Modulus of elasticity is a measurement of stiffness fora particular material. Every material has a different value for modulus of elasticity.The most common materials in temporary structures will be discussed in succeedingchapters. Modulus of elasticity is used in temporary structures to determine deflectionin beams and allowable buckling stress in columns. This will be discussed in moredetail in the following chapters.

Radius of Gyration Radius of gyration is a distance from the centroid toward theouter portions of the beam or column that resists buckling. Rectangles (lumber andtube steel) have a strong and a weak radius of gyration value when the x and y axesare different dimensions. Squares (lumber and tube steel) have the same radius ofgyration in both the x and y axis Circles have a radial radius of gyration; it’s the samefor 360∘. Finally, angle iron and similar sections have three axes of radius of gyration,and the weak axis for buckling is the z axis. This z-axis radius of gyration for anglesections should be used for buckling calculations if the designer wants to use theworst-case scenario. Figure 1.14 shows these three axes. The radius of gyration is a

FIGURE 1.14 Angle section showing three axes.

12 STATICS REVIEW

derivative of the cross-sectional area (A) of a section and its moment of inertia (I).Radius of gyration is calculated as follows:

r =√I∕A

where I = moment of inertia

A = cross-sectional area

Notice that the units work out to inches from the center of the cross sectionoutward. There is also a different radius of gyration for each axis, the strong andweak. Usually when column buckling is a concern, the weak r value is used becausethat is the direction that buckling will occur.

Example 1.4 What is the radius of gyration of a 4′′ diameter standard pipe with the

following properties?

A = 3.17 in2 and I = 7.23 in4√7.23 in4∕3.17 in2 = 1.51 in

FIGURE 1.15 Common shape properties.

1.3 STATICS 13

The radius of gyration for this 4′′ pipe determines how much buckling resistance

the pipe would have if used as a column or strut.

Figure 1.15 shows common shapes used in temporary structures and illustrates the

important properties.

Resultants The resultant of a force system is the system converted to a singleforce. The location of the resultant is determined by the shape of the system.Figure 1.16 shows examples of some simple resultants from a few different forcesystems.

The resultant of the rectangular force is located in the center.

The resultant of the triangular force is located one third the distance from the left.

The resultant of the trapezoidal force is located a distance (x) somewhere betweenthe triangular resultant and the rectangular resultant.

Statics students typically spend a great deal of time on force vectors, force trian-gles, and truss analysis. This text does not review these concepts as they are not usedwith basic temporary structure analyses. This will become evident in Chapter 3 whenthe different forces in temporary structures are introduced. In statics, it was also veryimportant to understand the difference between different types of supports: roller,fixed, hinged, and the like. Temporary structure supports are hinged the majorityof the time. The rest of the time, the supports or connections are fixed by weld-ing or bolting. The reason this is not common is because temporary structures are“temporary,” and any effort to fix connections cost money in labor and materials bothwhile installing and while removing them.

Free-body diagrams (FBDs) are drawings that summarize the forces (concen-trated), their directions, and the beam geometry and conditions. As learned in statics,the free-body diagram can be the whole structure or part of the structure. This texttends to draw FBDs of the whole structure. Those teaching temporary structuresshould always insist on a properly drawn FBD accompanying all temporary struc-tures computations. Mistakes most often occur when students have not taken time tocreate a complete FBD of the structure. A typical FBD is illustrated in Figure 1.17.

Moments Created by Forces A moment is a tendency of a force to createrotation. The point at which the moment is rotating is the center of the moment.

FIGURE 1.16 Resultant force location.

14 STATICS REVIEW

FIGURE 1.17 Free-body diagram example.

FIGURE 1.18 Static equilibrium.

The perpendicular distance between the force line of action and the center of the

moment is the moment arm. Therefore, a moment is the force times the moment arm

distance. If force is represented in pounds and the moment arm is represented by feet,

then the moment units are in foot-pounds (ft-lb).

Laws of Equilibrium This text focuses on static equilibrium. Static equilibrium

is when a body is at rest and all forces cancel all other forces out. To achieve static

equilibrium, three laws form the cornerstone to temporary and permanent design.

These are shown in Figure 1.18.

All forces in the horizontal direction must equal zero, ΣFx = 0.

All forces in the vertical direction must equal zero, ΣFy = 0.

All moments must equal zero, ΣM = 0.

Example 1.5 Determine the X and Y forces and the moment created in the simple

structure at support A with the forces shown in Figure 1.19.

FIGURE 1.19 Example 3 sketch.

1.3 STATICS 15

ΣFx = 0 −5 k + FVA = 0 FVA = 5 k

ΣFy = 0 +3 k – FHA = 0 FHA = −3 k

ΣM = 0 +(5k × 4 ft) − (3 k × 5 ft) = 0 MA = +5 ft-k

FVA = vertical force at point A

FHA = horizontal force at point A

MA = moment at point A

ΣFX = sum of forces horizontally

ΣFY = sum of forces vertically

Determining Reactions If we go back to Figure 1.1 and ask what are the reactionforces at the left support and the right support, this would be possible by adding themoments about each end of the beam.

Example 1.6 Determine the left and right reactions of the following beam where

P = 5 k, the beam is15 ft long, and the concentrated force is located 4 ft from the left

side as shown in Figure 1.20.

FIGURE 1.20 Simple beam reaction example.

To determine the reaction forces at R1 and R2, total the moments about one of the

two reactions. Let’s first add up the moments about R1.

ΣMR1 = 0 − [(5k)(4′)] + [(R2)(15′)] = 0

therefore, R2 = 1.33 k.

Now for the other reaction, we can use one of two methods. We can either total

the moments about R2, in order to get R1 (similar to what we just did), or we can add

up the forces in the vertical direction:

ΣMR1 = 0 +[(5k)(11′)] + [(R1V )(15′)] = 0

therefore, R1V = 3.67 k.

ΣFV = 0 −5 k + 1.33 k + R2V = 0

therefore, R2V = 3.67 k.

16 STATICS REVIEW

FIGURE 1.21 Simple beam resultant force solution.

It is recommended that the moment method by itself be used or that both methods

be used as a double check. The vertical force method can produce errors if the R1

calculation was incorrect. The engineer should also do a visual check. This means,

do the results make sense? For example, if we reversed the reactions by mistake and

did a visual check, we would realize that the larger reaction force is farther from

the 5-k force and that would not be logical. The larger reaction force has to be on

the side that the 5-k force is closest to. Figure 1.21 illustrates the resultant forces of

Example 1.6.

Example 1.7 Now let’s look at a more complicated example. Figure 1.22 contains

multiple loads and load types.

FIGURE 1.22 Simple beam with various loads.

This beam is 20 ft long; the 7-k (P1) load is 5 ft from the left side; the 5-k (P2)load is 9 ft from the left side; and the uniform load is 8 ft long, starts 12 ft from the

left, and extends to the far right of the beam.

Step 1: Since there is a uniform load, first it must be converted to a concentrated

load.

P3 = wL

where P3 = force in klf

w = uniform load

L = length of the beam

P3 = 2 klf × 8 ft = 16 k

Step 2: Place the P3 load of 16 k at the center of the uniform load and draw a

new FBD. Double check that the dimensions equal 20 ft. Draw an FBD

similar to Figure 1.23.

1.3 STATICS 17

P1=7 k

R1 R2

w

P2=5 k P3=16 k

5ʹ 4ʹ 7ʹ 4ʹ

FIGURE 1.23 Simple beam various loads resultant forces.

Step 3: Add the moments about the left reaction (R1) in order to calculate R2.

ΣMR1 = 0 −[(7k)(5′)] − [(5k)(9′)] − [(16k)(16′)] + [(R2)(20′)] = 0


Step 4: Determine R1 using both methods shown above in Example 1.1.

ΣMR2 = 0 +[(7k)(15′)] + [(5k)(11′)] + [(16k)(4′)] − [(R1)(20′)] = 0


ΣFV = 0 −7 k − 5 k − 16 k + 16.8 k + R1V = 0

therefore, R1V = 11.2k.

The visual test is not as easy this time, but with the 16-k load close to

the right support, it would be expected that the right side would have a

larger reaction.

CHAPTER 2

STRENGTH OF MATERIALS REVIEW

2.1 STRESS

When forces act on a structure, internal stresses are created that resist the forces over

a unit area. Materials experience several different types of stress depending on the

forces and loading conditions. The three most common types of stress are normal,

bending, and shear. In these three types, compression and tension play a large part on

how the stress occurs, especially in normal and bending stress. Stress is measured in

force over an area, such as pounds per square inch (psi), kips per square inch (ksi) or

pounds per square foot (psf). Generally, psi and ksi refer to steel and wood forces, and

psf designates soil and concrete pressure. Strength of a particular material is measured

by how that material can resist forces. When these forces are measured over a unit

area or section property, stress is the result.

2.1.1 Normal Stress

Normal stress is when a force is directly applied in compression or tension to a unit

area. The total force divided by the smallest cross-sectional area equals the highest

stress on the cross-sectional area. The higher the force and/or smaller the area will

result in a higher stress. On the other hand, a lower force and/or larger area will pro-

duce a lower stress. Normal stress is most commonly displayed using the following

formula, which can be modified to solve for any of the three variables as shown in

the other two formulas:

f = PA

P = fA

A = Pf

18

2.1 STRESS 19

where f = stress

P = axial force

A = cross-sectional area

2.1.2 Bending Stress

Bending stress occurs in beams that span from one support to another (or multiplesupports) and are loaded either uniformly, with concentrated loads, or both. When abeam is loaded, it begins to deflect. Deflection (to be covered later in this chapter)is acceptable as long as the beam is not overstressed and as long as the maximumdeflection requirement is not exceeded. The following formula calculates basic bend-ing stress. Section modulus is derived from moment of inertia, which was discussedin Chapter 1.

fb =MS

where fb = bending stress in the most extreme fiber of the member

M = maximum moment in the beam under load considering the span

length, the load, and the type of supports

S = section modulus of the beam section

2.1.3 Shear Stress

Shear stress occurs mostly where there are concentrated forces and reactions. Theforce at these locations is trying to fail the member, like scissors cutting paper. Oneside of the shear area is acting in the opposite direction of the force on the otherside of the shear area. The cross-sectional area of the member at this location is theproperty resisting this failure. The type of material is the other key component toresisting shear. Steel, wood, aluminum, and the like are going to have different shearresisting capabilities. Each of these will be discussed in this book under differentapplications.

For wood or any other rectangular sections, shear is calculated with the followingequation:

fv =1.5 VA

where V = maximum shear in beam

A = area of cross section

For steel sections, shear takes into account only the web sectional area and is cal-culated with the following equation:

fv =Vmax

twd


tw = web thickness

d = depth of beam

20 STRENGTH OF MATERIALS REVIEW

2.1.4 Horizontal Shear Stress

Horizontal shear occurs in the connections between flanges and webs of beams and

between the laminates in plywood. When a beam is placed in a bending condition,

it deflects the beam’s internal parts move horizontally. Without a solid connection

between the internal parts, these parts will slide horizontally. In order to resist this

sliding, one can use welding, bolting, gluing, and the like. In the case of a wide flange

beam, the flange and web are cast together monolithically. Therefore, the horizontal

shear resistance is substantial without additional connectors. The student of strength

and materials will figure out the weld size and length or the number and size of bolts

used in a composite connection. The National Design Standard (NDS) has established

horizontal shear stress values for plywood. This chapter will review the basic concept

behind horizontal shear. Horizontal shear is calculated by the following formula:

v = VQIb

where V = maximum shear in the beam, usually at the reaction point (pounds

or kips)

Q = Ad (in in3)d = distance from the centroid to the center of the portion of beam in

question (in)

A = cross-sectional area of the top flange (in2)

I = moment of inertia (in3)

b = width of the connection point from the flange to the web (in)

Example 1.2 is a good example to use to illustrate horizontal shear stress where

the top flange meets the web. If a wood flange was attached to a wood web, could the

horizontal shear be determined between the flange and the web? If this beam had to

span 20 ft with 200 pounds per linear foot (plf) throughout the length, what would be

the horizontal shear stress between the top and bottom flange? Figure 2.1 illustrates

this example, and Figure 2.2 shows the center of gravity.

Beam properties:

The thickness of the flange and the web are both 2′′. The flange is 6′′ wide (A).

The web is 8′′ tall, not including the flange thickness (B).

FIGURE 2.1 Composite T section.

2.1 STRESS 21

FIGURE 2.2 Center of gravity.

Cross-sectional areas of each member:

Area of A = 2 × 6 = 12 in2.

Area of B = 2 × 8 = 16 in2.

Distance from bottom to center of A = 8 + 1 = 9′′.

Distance from bottom to center of B = 1

2of 8′′ = 4′′(28)(y′) = (12)(9) + (16)(4);

y′ = 6.14′′.

In Chapter 1, the moment of inertia was calculated as I = 260.75 in4.

If this beam had to span 20 f𝛼 with 200 plf throughout the length, then the shear

at the ends would be

V = wl2

V = (200 lb∕ft)(20 ft)∕2 = 2000 lb

Area of the top flange is a = 12 in2

d = distance from the center of the top flange to the centroid of the composite as

calculated in Chapter 1 as 5.29′′ from the bottom. From the top, the centroid would be

10′′ − 6.14 = 3.86′′. Since the top flange is 2′′ thick, half of 2′′ would be the center

of the flange, therefore, 3.86′′ − 1′′ = 2.86′′ which is d.The second moment is the product of the area and d.

Q = Ad Q = (12 in2)(2.86 in) = 34.32 in3

b = thickness of the web where it connects to the top flange, b = 2 in.

Since all the values are now known, the horizontal shear can be determined.

v = VQIb

v = (2000 lb)(34.32 in3)(260.75 in4)(2′′)

= 131.6 psi

A quick check should be made with the units. The pounds remain and three of the

5-in units cancel, leaving inches squared in the denominator. Therefore, pounds per

square inch remain, which are the units of horizontal shear stress. If this shear stress

was compared to most allowable shear stresses on any specie of wood, this would be


considered high. The question remains if the flange and web can be attached with areasonable number of nails or an adhesive to resist the horizontal shear stress while

not damaging the web or flange.

2.1.5 Modulus of Elasticity

Modulus of elasticity for a material is the ratio between unit stress (psi) and the unitdeformation (in/in). The unit of measure remains pounds per square inch (psi). Theunit is symbolized with E and is an indication of stiffness of a material. Modulus ofelasticity is used in determining deflection in beams and buckling in columns.

For steel, E is approximately 29,000,000 psi. For wood, E is between 1,300,000and 1,900,000 psi, depending on the specie of wood. For concrete, E ranges from2,000,000 to 5,000,000 psi depending on the mix design. This text does not study

concrete design.

2.2 BENDING MOMENTS

2.2.1 Maximum Bending Moments

Shear and moment diagrams are used when these values cannot be calculated fromour simple, standard formulas such as wL2∕8, PL∕4, or Pa and when our support

arrangement is something other than a simple support. Construction managementand engineering students spend a great deal of time learning how different beamsreact under certain loads. Rather than detailing all aspects of beam design, this bookwill review the overall concepts. Later, shear and moment diagrams will be discussed

in detail. Figure 2.3 shows the three most common and simple maximum momentformulas used: from left to right, uniform load, concentrated load in the center ofspan, and concentrated load off center of span, respectively. If these arrangements

are encountered, shear and moment diagrams are not necessary.Chapter 3 will discuss different loading conditions and where the loads might

come from in a temporary structure condition. For the purpose of this book, the loadson beams are classified as either concentrated or distributed. In temporary structure

design, both are used. However, the distributed load is probably more common. Steelbeams that are supported by columns typically support other beams. If the beamsbeing supported are close together, then this may be analyzed as a distributed load.

If these same beams are spaced more than one fourth the length, they may be consid-ered as individual concentrated loads. Moments are created in beams when they aresubjected to bending.

The load on an individual beam is the concentration of forces that are within the

beam’s tributary width. The tributary width of a beam is the distance that extends

FIGURE 2.3 Load types.

2.2 BENDING MOMENTS 23

FIGURE 2.4 Tributary width on a beam.

halfway to the beam on each side, which is also equal to the spacing of the supporting

beams. If beams are spaced at 4 ft on center, one beam sees the forces from 2 ft each

side or 4 ft of tributary width. Figure 2.4 illustrates the tributary width on one beam,

spaced equally.

2.2.2 Maximum Shear

Beams are typically supported by columns or other beams. These points of support

are reaction points and points at which high shear value occurs. At the reaction points,

the beam has a concentrated shear force where the reaction force is directed upward

and the beam load is directed downward at the same point. A shear diagram is used

to show the values of shear at any given point on a beam. The shear value can be

calculated at any point along a beam from one end to the other and is dependent on

the types of loads, the support locations, and types of supports. In this book, most

supports will be free or hinged. It is not common to fix (weld or bolt) connections of

beams and columns in temporary structures because it adds unnecessary costs to the

installation and removal of the elements.

The magnitude of the shear at any point of a beam is equal to the sum of all the

forces to one side (usually to the left). These forces have positive and negative values.

Forces directed upward (reactions) are positive and forces directed downward (con-

centrated loads and distributed loads) are positive values. The difference between the

two is the amount of shear at a particular section.

2.2.3 Law of Superposition

Law of superposition comes in handy when more than one common force is applied

to a simple span, but it is not quite complicated enough for a shear and moment dia-

gram. For instance, if a simply supported beam was loaded with a uniform load and

a single or multiple concentrated load, the maximum moment could be calculated by

adding the moment caused by the uniform load to the moment caused by the con-

centrated load. The combination of concentrated loads either fall into the condition

of Figure 2.3(b) or 2.3(c). Figure 2.5 illustrates three concentrated loads on a simply

supported beam.

The law of superposition says that the moment of inertia for the beam and forces

shown can be calculated by adding the maximum moment formula of the P1a to the

maximum moment formula P1L∕4. This, of course, only works if the two end forcesare the same distance from the supports and the center force is in the middle of the

beam:

Mmax = P1a +P1L

4


FIGURE 2.5 Law of superposition.

2.3 MATERIALS

This book will focus on the two most common materials used in construction: steeland wood. Steel is a manufactured product that results from its chemical compositionbeing cast during a heating process. For steel, its carbon content is the component thathas the greatest effect on its strength. The steel type used in most temporary structuresis ASTMA36, the number 36 representing the steel’s yield stress point during Ameri-can Society for Testing and Materials (ASTM) testing. Unless specifically noted, thistype will be used for examples in this book.

2.3.1 Factors of Safety

Factors of safety (FOS) are reductions applied to material stress values that lower theamount of stress a material can see in its useful life. For this book, the useful life ofa material has been defined as several hours to several months and it is consideredtemporary. Factors of safety can be applied to yield stresses or ultimate stresses of aparticular material. For the purpose of discussion, mild A36 steel will be used. Thereare three points of stress we will use to understand factor of safety: (1) the ultimate(Fult), (2) the yield (Fy), and (3) the allowable (Fall). The yield stress of a materialis typically close to the plastic limit. The yield is the point at which a material, suchas steel, begins to deform, but when the load is released will go back to its originalshape. The ultimate stress of a material is achieved when a material fails. If we usemild A36 steel as an example, we can understand these points of stress.

2.3.2 Grades of Steel

Steel comes in many grades for both structural shapes and bolts. In construction, themost common steel is A36 (grade 36). The 36 represents the steels yield strengthof Fy = 36 ksi. The number system used in grade of steel is named after the ASTMtesting methods. A36 steel is considered mild steel because of its carbon content.As the carbon content changes, the steel strength and corrosion resistance changes.Other steel grades used in construction are A328, A572 (grade 50, 50 ksi), A572(grade 55, 55 ksi), andA572 (grade 60, 60 ksi). There aremany other grades available.As the grade increases, typically so does the price per pound. It is becoming morepopular for an engineer to specify a higher grade steel in order to maximize the designand, hopefully, still be economical.

An easy and common sample would be a 1′′ × 1′′ steel bar placed in a labora-tory testing machine. The specimen would be pulled in tension and the forces would

2.3 MATERIALS 25

FIGURE 2.6 Stress–strain graph.

be recorded. The steel would immediately begin to undergo stress (pounds per squareinch, psi) as the force is increased. This applied force will continue to increase alongwith the stress as shown in Figure 2.6. This is a linear function until a certain point.When the force reaches approximately 32,000 lb, the steel sample begins to deform.This deformation is in the form of elongation. If the force is removed at this point,the sample would go back to its original length. This is the plastic limit. When theforce is increased beyond 36,000 lb, the elongation continues. Beyond this point, ifthe load is removed, the sample would maintain a portion, if not all, of the deforma-tion. As the force is increased even further, the sample continues to elongate. Whenthe force reaches approximately 58,000–70,000 lb, it will see its ultimate stress andfail. Ductility is a material’s ability to deform in tension and stretch to a longer length.

Some key data points during this test of stress are:

Yield stress (Fy) 36,000 lb∕1 in2 = 36,000 psi

Ultimate stress (Fu) 58,000 lb∕1 in2 = 58,000 psi

The value derived from these data points is that allowable stress (Fb) 21,600–24,120 psi (depends on 0.6−0.67 FOS).

Generally, a factor of safety is applied to the yield stress. In other words, the allow-able stress (Fb) is calculated by factoring down the yield stress. If an engineer wantedto apply a 1.5 : 1 factor of safety, he would divide 36,000 psi by 1.5. In this case theallowable stress (the maximum amount of stress the steel could withstand withoutfailing) would be 24,000 psi. If the engineer followed the recommendation of theAmerican Institute of Steel Construction (AISC) and applied a 60% factor of safety,the allowable stress would be 36,000 psi × 0.60, or 21,600 psi. A 67% factor of safetywould multiply 36 ksi by 0.67, or 24.12 ksi. This example for steel would representallowable “normal stresses.” Other types of stress such as shear and normal stresswould have a lower factor of safety. Table 2.1 shows the most common types of stressfor materials used as temporary structures.

2.3.3 Compact Beam

The factor of safety can be adjusted to 0.66Fy if it meets a compact section require-ment. To see if a beam meets this requirement, determine its ratio of d/t. This is theratio between the beam’s depth and minimum thickness of its vertical component(web):

dt<

√190

Fy

√190

36 ksi= 31.67


TABLE 2.1 Allowable Stresses for A36 Steel

Stress Type Factor of Safety Allowable Stress (psi)

Fb (bending) 0.6–0.67Fy 21,600–24,120

Fv (shear) 0.40Fy 14,400

Fc (compression) 0.90Fy 32,400

Ft (tension) 0.60Fy 22,000

0.50Fu 29,000

0.33Fu 19,000

E Modulus of elasticity 29,000,000–30,000,000 psi

TABLE 2.2 Compact Beam Sample Calculations

Beam d (in) T (in) d∕t (ratio) Conclusion

C6 × 8.2 channel 6.0 0.20 30.0 30 < 31.67, compact

W12 × 58 WF beam 12.19 0.36 33.86 33.86 > 31.67, not compact

TS8 × 8 × 1∕2 8.0 0.5 16.0 16 < 31.67, compact

where d = depth

t = web (vertical) thickness

Fy = yield stress or 36 ksi

Table 2.2 illustrates calculations that check whether beams meet the compact

requirements.

When a beam is considered compact, Fb = 0.67Fy.

2.3.4 Wood

Plywood Plywood has a strong and weak direction, depending on the direction of

the supporting members (studs or joists). This will be discussed in detail in Chapter 9,

but Table 2.3 has been provided to show the allowable stresses of standard grade

BB-OES plywood used. It is common for contractors to use other grades of plywood,

typically higher grades such as high density overlay (HDO) or medium density over-

lay (MDO). The manufacturer of these grades of plywood will supply a property data

sheet including allowable stresses, deflection, spans, and so forth.

Dimensional Lumber Dimensional lumber is the most popular material used for

formwork and other lightweight temporary structure systems. Its versatility, weight,

TABLE 2.3 Plywood Allowable Stress Ranges

Strong (Face Grain Parallel to Span)

Bending stress (Fb) 1500–2000 psi

Shear stress [Fv(Ib∕Q)] 55–60 psi

Modulus of elasticity (E) 1,500,000–1,600,000 psi

2.4 DEFLECTION 27

TABLE 2.4 Dimensional Lumber Allowable Stress Ranges

Strong (Face Grain Parallel to Span)

Bending stress (Fb) 700–2400 psi

Shear stress (Fv) 85–145 psi

Adjusted shear stress (F′v) 170–190 psi

Compression, parallel to grain (Fcll) 400–730 psi

Compression, perpendicular to grain (Fc) 625–1650 psi

Modulus of elasticity (E) 1,200,000–1,900,000 psi

TABLE 2.5 Load Factor Uses

Factor Permanent (P) or Temporary (T)

Duration (permanent to impact) T

Wet use (moisture) P, T

Flat use (loaded in weak direction) P

Size (timber vs. lumber) P

Temperature (heat and cold) P

Beam stability (lateral stability) P

Shear T

Incising (pressure treatment) P

Repetitive member (multiple uses) P, T

Column stability (slenderness) P

Buckling stiffness (slenderness) P

cost, and ease of use make it a highly preferred material for temporary structures.Table 2.4 shows the ranges of strengths in dimensional lumber.

Load factors are multipliers that adjust the allowable stresses either up or down,depending on the factor. For permanent design, there are approximately 10 factors.Temporary structures use only a few of these, such as load duration, wet use, shear,and temperature. The range depends on the type of use and severity of each. Forinstance, temperature factor will depend on how extreme the temperature goes upand down. The wet use factor will depend on the variability of moisture. The fac-tors only affect specific stresses. One factor may affect bending stress and modulusof elasticity, whereas another factor may affect shear and compression parallel tograin. For example, compression perpendicular to grain requires only the wet useand temperature factor. Some of these available factors are shown in Table 2.5.

2.4 DEFLECTION

Deflection occurs in all loaded beams. It is defined as the measurement between thestraight, undeflected beam and the deflected beam, typically in inches. Some beams,if their spans are long enough, deflect under their own weight as well. In this case,during the beam analysis, the weight of the beam should be added as an additionaluniform load.

Deflection is compared to engineering standards such as L/240, L/360, L/480,and so forth. These standards determine the maximum deflection allowed per span


of beam. For example if a 32-ft beam span had to stay within an L/360 standard, the

maximum deflection would be

32 ft × 12′′∕ft∕360 = 1.0667 in

Note that L is converted to inches in the numerator.The amount of deflection seems relatively high; however, for a beam to deflect

1.0667 in. in 32 ft is negligible when it comes to temporary structures if the movement

in this beam does not affect any other part of the system. In permanent construction,the excessive deflection of a stud or joist may cause the stucco or drywall to crack;

therefore, the deflection limits will take this into consideration and be more stringent,such as L/480. Also, the stress on this same beam with this same deflection wouldtypically be low. Generally, in temporary beam design, the stress on the steel dictates

over the amount of deflection within a standard factor of safety of 1.5 or 2.0 : 1.0.Deflection tends to govern only when there is a light load over a very long span.

When beams are designed to resist bending stress and they have an appreciable

amount of deflection, the designer would typically calculate the amount of deflectionand add camber strips to the top of the beam to offset the deflection anticipated. Thisway, when the beam deflects, the top of the beam becomes a straight line again, thus

not leaving undulations in the finish product. Chapter 11 discusses the use of camberstrips in bridge construction. A very common deflection calculation for a uniform

load on a simply supported beam is

5wl4

384EI

In temporary structure design, L/360 is very common. A lower tolerance couldbe encountered in a support of excavation system where horizontal movement thatcauses some settlement adjacent to the excavation is acceptable due to the lack of

structures nearby. However, in another case, the requirements for a temporary struc-ture may be more stringent. For example, the engineer designing excavation supportwithin the city streets of San Francisco may call for an L/1000 tolerance on the

shoring design so the displacement of the adjacent roads would be immeasurable.The shoring system selected, deep soil mixing with wide flange beams embedded or

sheet piling with waler support beams, is designed for almost no lateral movement.A similar tolerance was required on the SR 520 Evergreen Floating Bridge Pon-

toons in the State ofWashington. The formwork that was used had steel plate sheeting

instead of plywood in order to meet the engineer’s deflection requirements. Deflec-tion requirements such as these examples were not frequently encountered until morerecently.

2.5 SHEAR AND MOMENT DIAGRAMS

Shear and moment diagrams are a staple to engineering temporary structures designwhen the loading and support conditions are less common. As mentioned before,

there are three simply supported loading conditions where the maximummoment canbe determined through a simple formula. In addition, if more than one of these three

2.5 SHEAR AND MOMENT DIAGRAMS 29

conditions are combined, the law of superposition can be used. As a rule of thumbwhile creating shear and moment diagrams, clockwise rotation will be a negativemoment and counterclockwise rotation will be a positive moment. Positive shear willbe upward and negative shear will be downward.

When the standard formulas cannot be used, two options can be used. The first isto create a shear and moment diagram, which we will do in this section. The secondwould be to use beam software that calculates these values from some simple inputdata provided by the temporary support designer. This software will be covered laterin this chapter and will be used for structures classified as indeterminate structures,those having more than two unknown reactions.

In statics, strength of materials, or mechanics classes, students spend a great dealof time mastering shear and moment diagrams. This text will refresh the user onthe basics of shear and moment diagrams without repeating the methods for exactvalues at specific locations along the beam. The methods used here will only derivethe maximum and minimum values for shear and moment. Basic shear and momentdiagrams done manually include the following:

1. Two supports (two unknowns). More than two unknowns are indeterminatestructures, which are covered later in this chapter.

2. Concentrated loads, uniform loads, or varying uniform loads.

The method used in this text builds the moment diagrams from the areas of theshear diagrams. It is very important to build the shear diagrams correctly so themoment diagrams are accurate. Basically, the sum of the areas to the left of the sheardiagram at any point is equal to the moment in the beam at that point. Following arethree examples of building shear and moment diagrams. The first will be a very sim-ple condition showing how M = wl2∕8 is derived; the second is a relatively simpleexample with two loads; and the third will be intentionally complicated (although notvery realistic) in order to assure that the concepts are mastered.

Example 2.1 Draw a shear and moment diagram for the following loading condi-

tion: simply supported beam (l), uniform load (w) the same length as the beam.

Step 1: Determine the (R)esultant of the uniform load and its location along the

beam.

R = wl, distance1

2centered on the beam

Step 2: Determine the shear/reactions at the left and right side supports:

Rl = Rr = wl∕2

Step 3: Draw the shear diagram with these reactions. The diagram has a positive

shear at the left equal to the reaction at the left (+wl∕2). The shear valuedecreases as the shear decreases from left to right. The amount of decrease

is equal to the value of w per linear measure from left to right. When the

uniform load w gets to the right support, the shear value should be −wl∕2and come back to a zero value.


FIGURE 2.7 Shear and moment diagram for Example 2.1.

Step 4: Draw the moment diagram by determining the points of change from the

shear diagram and the area of the shear at different locations along the

beam. Since this example is just a single uniform load, there are no abrupt

changes in shear except at the end reactions, and it only crosses zero shear

once. The zero shear locations are an indication of a maximum point of

moment, either positive or negative. Figure 2.7 represents the shear and

moment diagram for Example 2.1.

It should be pointed out that the shear diagram is broken into two halves, both

of which are triangular shapes. The maximum moment is in the center of the beam

(where shear value crosses a zero), and its value is equal to the area of the shear

diagram from the center to the left support:

M = Area = bh2

=(L∕2)(wL∕2)

2= wL2

8

This same process can be used to prove m = PL∕4 and Pa. Have your instructorillustrate these in class or try this during a study session.

Example 2.2 Draw a shear and moment diagram for the following loading condi-

tion: simply supported beam L = 20′, uniform load of 5 kips per linear foot (klf) the

same length as the beam length and a concentrated load of 25 k located 8 ft from the

left support. This is represented in Figure 2.8.

Step 1: Determine the resultant (R) of the uniform load and its location along

the beam.

5 klf × 20 ft = 100 k, located in the center of the 20′ span. Add this tothe 25-k concentrated load, and we have a total resultant downward force

of 125 k. This is the value that the reactions should equal in the next step.

Step 2: Determine the shear/reactions at the left and right side supports by adding

the moments about one side or the other. Let’s use the left side.

ΣMleft = 0 −(25 k × 8 ft) − (100 k × 10 ft) + (Rr × 20 ft) = 0

therefore Rr = 60 k.


FIGURE 2.8 Loading diagram for Example 2.2.

Now add the vertical forces, +60 k − 125 k + Rl = 0; therefore, Rl =65 k.

Step 3: Draw the shear diagram with these reactions. The diagram has a positive

shear at the left equal to the reaction at the left +65 k. The shear value

decreases as the shear decreases from left to right. The amount of decrease

is equal to the value of w per linear measure from left to right. The amount

of shear loss due to the uniform load is 5 klf × 8 ft = 40 k. Now the shear

is at a positive 25 k (65 k − 40 k = +25 k). When the uniform load (5 klf)

gets to the 25-k concentrated load (8 ft from the left support), there is an

abrupt change in shear equal to the 25-k load. Here the shear diagram

drops straight down 25 k, the shear value is now 25 k less or 0 k (25 k −25 k = 0 k). From here the shear diagram continues the downward slope

due to the uniform load of 5 klf. There are 12 ft to go, therefore 5 klf ×12 ft = 60 k. Subtract this 60 k from the 0-k shear and this equals −60 k.

Now, finally go up the amount of the reaction at the right side (60 k), and

this is the end of the shear diagram. We are at 0 k of shear at the right

support, the same place we started at on the left side. Figure 2.9 shows the

diagram for this example.


Step 4: Draw the moment diagram by determining the points of change from the

shear diagram and the area of the shear at different locations along the

beam. The most significant location is at the 25-k concentrated load 8 ft

from the left side. Also notice the shear diagrams are triangles and trape-

zoids. Therefore to calculate the areas under the shear diagrams, one must

separate the triangle from the rectangle. Zero shear is always an indica-

tion of a maximum point of moment. The most important values when

determining the moments are the maximum values, so the shape of the

parabolic curve is not as important as the value at the extreme points.

After all, to determine the minimum beam size in beam design, one must

know the maximum moment. In this case the maximum moment is equal


to the area within the shear diagram. The positive shear and the negative

shear will equal each other in order for the shear diagram to equal zero

value at each end.

The shear diagram is a trapezoid (triangle and a rectangle); therefore, the area of

both need to be calculated in order to come up with the total moment.

Top triangle (A1)∶ A1 =40 k × 8′

2= 160 ft-k

Bottom rectangle (A2)∶ A2 = 25 k × 8′ = 200 ft-k

Total area∶ Mmax = 360 ft-k = max moment

Figure 2.10 illustrates this calculation.

FIGURE 2.10 Maximum moment.

Example 2.3 Draw a shear and moment diagram for the following varying linear

load with a max of 500 plf on a 12-ft beam with supports at the left and 4 ft from the

right side. Figure 2.11 depicts the diagram for Example 2.3.

FIGURE 2.11 Loading diagram for Example 2.3.

Today, this condition would be input into a beam software program and the results

would be seconds away. However, in order to get confident in producing an accurate

shear and moment diagram, this solution will be derived manually.

Step 1: Calculate and locate the resultant of the triangular linear load.

(500 − 0)(12′)∕2 = 3000 lb located one third from the left and two

thirds from the right.

Step 2: Determine the reactions at both supports, by summing the moments about

one side or both:

Rl = 1500 lb

Rr = 1500 lb

It is a coincidence that they are equal. This is only because of the trian-

gular linear load over a beam with a second support two thirds to the left.


Step 3: Draw an FBD using the beam length, support locations, and the resul-

tant as a concentrated force. Figure 2.12 shows the free-body diagram for

Example 2.3.

FIGURE 2.12 Free-body diagram for Example 2.3.

Step 4: Draw the shear diagram and moment diagrams. This is shown in

Figure 2.13.


Example 2.4 Beam Software ProgramsThere are an unlimited number of beam software programs available on the market

today. Some are much easier than others. It is recommended that the most basic and

easy-to-use software available be used because even the most elementary program

typically does more than the temporary structure designer needs. The program should

be able to do the following:

1. Accept multiple supports, hinged or fixed.

2. Accept uniform, concentrated, or varying uniform loads that represent all con-

struction loads.

3. Accept beam properties such as moment of inertial values.

The software should be able to produce the following:

1. Shear diagram with support reactions

2. Moment diagram indicating positive and negative moments and identifying

which one has the highest magnitude

3. Deflection diagram (optional)

Consider an example of a useful diagram produced by simple beam software.

The diagram is duplicating a typical formwork waler as the beam. The supports are


FIGURE 2.14 Software-generated S and M diagram.

the ties; and since there are more than two, this is already a complicated shear and

moment diagram because it is indeterminate. Finally, the loading is indicative of a

concrete pressure diagram, which is typically rectangular until it gets close to the

top and then tapers down to 0 SF. The program calculates the reactions at each tie

(support), the shear diagram, the moment diagram, and a deflection diagram. The

deflection is only accurate if a beam was selected while setting up the beam infor-

mation. However, usually the purpose of the diagram is to determine the maximum

moment before selecting the beam. Therefore, sometimes the deflection is not accu-

rate until the beam is selected, the steel type is selected, and the moment of inertia has

been input into the program. Figure 2.14 shows a typical software-generated diagram.

2.6 BEAM DESIGN

As mentioned earlier when discussing steel beams, the maximum moment of a beamunder a particular load and support arrangement is the most important information inorder to determine the beam size necessary to resist the bending stress in the beam.For wood beams, this information with the Shear resistance is combined to designthe beam. The information from the shear diagram is helpful for this check. Thesetwo checks, however, do not guarantee the beam will not have stability problems.Stability will be covered later in two ways. Early on, a ratio between the depth andthe flange width will be used as a quick way to check stability. The second method,to be covered later in this book, will be a more accurate measure of a beam’s stabilityusing the ratio d∕Af .

2.6 BEAM DESIGN 35

For this chapter, beam design will only have to comply with bending stress and

shear stress. It will become very apparent that shear in steel beams is rarely, if ever, anissue. For wood, however, shear can definitely be the governing criteria of the beamselection. Once again, here are the bending moment and shear formulas for both steel

and wood.

Steel: fb =MS

where S = section modulus of the beams horizontal axis

M = maximum moment

fv =Vdtw


tw = web thickness

d = depth

Wood: fb =MS

where S = section modulus of the beams horizontal axis

M = maximum moment

fv =1.5VA


A = cross-sectional area of the beam (A = bd)b = base dimension of wood beam

d = depth of wood beam

In Example 2.2, a beam is supporting a concentrated load and a uniform load.These values will be used for the steel beam design. The maximum shear and moment

were as follows:Vmax = 65 k

Mmax = 360 ft-k (360,000 ft-lb)

Steel Beam Design Before a beam can be designed, the allowable bending andshear stress should be determined. Using AISC standard values for stable beams,allowable bending stress (Fb) should not exceed 0.667Fy = 24 ksi and allowable

shear stress (Fv) should not exceed 0.40Fy = 14.4 ksi. The steel beam will bedesigned for bending stress and then checked for shear stress.

Step 1: Bending Stress—Determine the minimum section modulus using the

allowable bending stress and maximum moment in the beam from above.

Smin = 360 ft-k × 12′′∕ft∕24 ksi = 180 in3

List wide flange beams between 12′′ and 24′′ with a minimum section modulus of180 in3. Table 2.6 shows the available beams.


TABLE 2.6 Available WF Beams Meeting

Required Section Modulus Requirement

Beam Sx (in3)

W12 × 136 186

W14 × 120 190

W16 × N/A

W18 × 97 188

W21 × 93 192

W24 × 84 196

Looking at this list of beams, we can only make a decision based on the weight

of the beam and the fact that they all have an acceptable section modulus. Based on

weight (which is how steel is purchased, by the pound), the W24 would be the most

economical beam. Without any other criteria to select the beam by, such as stability,

weight will be the determining factor. Later in this book, stability of beams will be

addressed.

Depending on the number of beams a project would need in this example, cost

savings could amount to a great deal of money when comparing two of the beams

above. For instance, assume that the project required 40 of these 15-ft beams and the

staff was deciding between the W24 × 84 and the W18 × 97. The weight difference

per linear foot between the two beams is 13 lb (97 − 84 = 13). The total linear footageof beams required is 40 each × 15 ft = 600 lf. At 13 lb/lf, the difference in cost would

be 600 lf × 13 lb∕lf = 7800 lb of steel. If the cost of steel was $0.65/lb, this would

equal $5070.00.

Wood Beam Design

Example 2.5 Example 2.4 has loads and a span that generate amuch higher moment

than a normal size wood beam would be able to handle. Therefore, another beam

condition will be developed in order to explain wood design. Figure 2.15 illustrates

a 15-ft span simply supported beam with a 2000 lb/lf load from left to right (similar

to Figure 2.14).

Using the simple moment formula for a simply supported beam with a uniform

load:

Vmax =(2000)(15)

2= 15,000 lb

Mmax =(2000)(15)2

8= 56,250 ft-lb

Before a wood beam can be selected, the specie of wood should be known or at

least the allowable stress values. In this example, it will be assumed that the allowable

shear stress of the wood is Fv = 140 psi, and the allowable bending stress will be

Fb = 1500 psi. Using the formulas above, the beam size can be determined using

both bending and shear. The beam should be sized for bending and then checked

for shear.

2.6 BEAM DESIGN 37

FIGURE 2.15 Example 2.5 shear and moment diagram.

Step 1: Determine beam size based on bending:

1500 psi =(56,250 ft-lb)(12′′∕ft)

Sx

where Sx = 450 in3.

Table 2.7 suggests that the 6 × 24 (Sx = 506.2 in3) is the most econom-

ical because it has a minimum section modulus greater than 450 in3 while

maintaining the lowest cross-sectional area. Cross-sectional area deter-

mines the cost relating to board feet. Therefore, if the board has less board

feet (bf) per linear feet (lf), then it will be cheaper.

Step 2: Shear Stress—The shear stress should be checked before determining this

to be the best beam. The maximum allowable shear stress (Fv) from above

is 140 psi.

140 psi = 1.5 (15,000 lb)A

and A = 1.5(15,000)140 psi

= 160.7 in2

The area of a 6 × 24 is 129.3 in2, therefore Aall = 160.7 > 129.3

(Not Good).

The area of the 6 × 24 is not sufficient. Another way to check this

would be to solve for fv and compare to 140 psi:

fv =1.5 (15,000 lb)

129.3 in2fv = 174 psi > 140 psi (Not Good)

Since this beam fails the shear test, the next beam that met the Sx test(Sx ≥ 450 in3) will have to be tried.

fv =1.5(15,000 lb)

166.3 in2fv = 135.3 psi < 140 psi (OK)


The 10 × 18 works for bending and shear. A table containing values

for wood properties can be found in Appendix 6.

Cost Comparison for Wood Beams The cost of wood can also be made part

of the wood example. How does the cost compare between the 6 × 24 and a 10 × 18,

which had an adequate section modulus of 484.9 in3 (greater than 450 in3)? The 10 ×18 has a cross-sectional area of 166.3 in2 and the 6 × 24 has a cross-sectional area of

129.3 in2. For this example, the price of timber is assumed to be $1.20/bf. Table 2.7

provides a cost comparison of two sizes of lumber.

Similar to the example for the steel beam design, the cost savings can be very

significant. For example, the 600 linear feet used earlier would equate to

600 lf × ($18.00 − $14.40) = $2160.00 cost difference

Buckling Stress When explaining buckling stress, the terms “post,” “column,”

and “strut” are used interchangeably. When posts, columns, and struts are supporting

an axial force, the tendency is to buckle in their weak direction. The weak direction is

the direction parallel to the smaller dimension of the post. Buckling is basically like

deflection in a beam but is caused by an axial force rather than a bending moment.

If a post is square or if a pipe is used as a post, the post can buckle any direction

of its 360∘ (the square in 90∘ directions). If a rectangular post is used, then the post

will deflect in the weak direction as shown, unless appropriately braced. Figure 2.16

provides a sketch of column buckling.

Buckling stress is the amount of stress a post, column, or strut can withstand con-

sidering (1) its weakest radius of gyration, ry, and (2) its largest unsupported length,

lu. Normal stress of an axial loaded post, when buckling is not considered, is P/A.For A36 steel, the allowable would be approximately 27,000 psi. For wood posts, the

allowable would be in the range of 625–1650 psi. However, when buckling is consid-

ered, taking into account ry and lu, the allowable stresses (buckling stress) can drop aslow as 3000 psi (estimate) for A36 steel and 150 psi (estimate) for different species of

wood. These numbers are ballpark figures in order to make the point. These changes

TABLE 2.7 Cost Comparison of Lumber

Beam Area (in2) bf per lf Cost per bf ($) Cost per lf ($)

6 × 24 129.3 12 1.2 14.40

10 × 18 166.3 15 1.2 18.00

FIGURE 2.16 Column buckling sketch.

2.6 BEAM DESIGN 39

in allowable stress values can be considered adjustments to the P/A, axial normal

stress value.

Slenderness ratio (SR) is the ratio between the length of the post in inches and the

weakest radius of gyration or the smallest dimension of a rectangle, depending on

whether it is steel or wood. The ratio of an 8-ft long 2 × 4 S4S would be l∕d (8′ ×12′′∕ft)∕1.5′′ = 64. This is a ratio because the inches cancel so there are no units. The

ratio, if we lengthened the beam to 12 ft, would be 96 (50% higher). The higher the

slenderness ratio, the more susceptible the section (2 × 4 in this case) is to buckling.

Instead of P/A being 1250 psi, the allowable buckling P/A would be below 100 psi.

This is a considerable difference when a beam is being used as a post, column, or strut.

In strengths of materials, the student studied the Johnson and Euler theories of

column buckling. These two theories distinguished between long and short columns,

depending on their slenderness ratios being above or below 120. This text uses one

method to calculate either scenario. A table will also be introduced in Appendix 4 for

trial-and-error column design calculations.

End connections of a column are crucial to the proper design. The effective length

is the adjusted length after taking into account the effective length factor, which is

determined by how the ends of the column are attached. The ends are typically pinned

in temporary structures; however, other options are fixed (welded or bolted), free like

a flag pole, and fixed but free to move laterally. Table 2.8 shows the multiplying factor

used to adjust the effective length for the slenderness ratio calculation.

Two formula methods for columns will be used: one for steel and one for wood

members. The slenderness ratio will be used for both; however, steel will use the

weakest radius of gyration, while wood will use the least dimension in cross section.

Steel:

Fbs = 𝜋2E[(klu∕r)]2 × FOS

= PA

(units in psi or ksi)

where Fbs = allowable buckling stress (psi)

E = modulus of elasticity for steel, usually 29,000,000 psi

k = connection-type multiplier, depending on fixed, hinged, etc. (no units)

lu = unsupported length of column; if there are supports, then the longest

distance between supports (in)

r = weakest radius of gyration (in)

P∕A = normal stress and is set equal to allowable buckling stress in order to

determine

Pall = (or area allowed)

Refer to Appendix 2 for pipe properties.

TABLE 2.8 End Connection Factors

Connection Type Factor

Pinned × Pinned 1.0

Fixed × Pinned 0.7

Fixed × Fixed 0.5

Fixed × Free 2.0


Example 2.6 Buckling of a Steel ColumnLet’s look at a 16-ft-long 3

1

2′′ standard steel pipe with an FOS of 1.5 : 1.0. To deter-

mine the allowable axial load this pipe can support, follow these steps.

Step 1: Look up the cross-sectional area (A) and the radius of gyration (r). Thereis only one value for r because it is a pipe.

A = 2.68 in2

r = 1.34 in

Step 2: Calculate Fbs = 𝜋2(29,000,000)∕{[(1.0)(16 × 12′′∕ft)∕1.34′′)]2 × 1.5} =9294 psi.

Step 3: Set Fbs = Pall∕A and solve for Pall = 9294 psi × 2.68 in2 = 24,909 lb.

This 16-ft-long, 31

2′′ standard pipe can support 24,909 lb before failing by buck-

ling. If buckling was not an issue, then Pall = 27,000 psi × 2.68 in2 = 72,260 lb. This

is a 47,451-lb reduction because of buckling.

Wood Columns:Fbs = 0.30E∕(lu∕d)2


E = modulus of elasticity for the wood specie (psi)

lu = unsupported length of column if there are supports, then the longest

distance between supports (in)

d = least dimension of the cross section

Table 2.9 explains how d is the least dimension of a cross section. A square cross

section has the same dimension for d, whereas, a rectangular cross section only has

one d, the smaller one.

TABLE 2.9 d Dimensions for

Various Lumber Sections

Size d Dimension (in)

2 × 4 1.5

4 × 4 3.5

4 × 6 3.5

4 × 8 3.5

6 × 6 5.5

The FOS safety for wood is built into themodulus of elasticity value. P/A is normal

stress and is set equal to allowable buckling stress in order to determine Pall or areaallowed.

Example 2.7 Buckling of a Wood ColumnLet’s look at a 12-ft-long, 4 × 4 with E = 1,700,000 psi. To determine the allowable

axial load this 4 × 4 can support, follow these steps.

2.6 BEAM DESIGN 41

Step 1: Look up the cross-sectional area (A) and the radius of gyration (r). Thereis only one value for r because it is a pipe.

A = 3.5 × 3.5 = 12.25 in2

d = 3.5 in

Step 2: Calculate Fbs = 0.30 (1,700,000 psi)∕[(12 ft × 12′′∕ft)∕(3.5)]2 = 301.3

psi.

Step 3: Pall = 301.3 psi × 12.25 in2 = 3690.3 lb.

This 12-ft-long, 4 × 4 can support 3690.3 lb before failing in buckling. If buckling

was not an issue, then Pall = (650 psi − 1650 psi) × 12.25 in2 = 7963 lb to 20,213 lb

depending on the specie of wood. This is a 4000 lb − 17,000 lb reduction because of

buckling.

Example 2.8 Steel Column Design Using HP ShapesDetermine the allowable load (Pall) of a HP 14 × 89 that is 32 ft long. Appendix 3

will be used for this example.

Step 1: Look up the properties of a HP14 × 89 in Appendix 1.

ry = 3.53 in

A = 26.1 in2

Step 2: Calculate the slenderness ratio (SR).

SR =32′ × 12′′∕ft

3.53 in= 108.8

Step 3: Look up the allowable buckling stress of SR = 108.8 in Appendix 4 for

36 ksi steel.

Fbs = 11.91 ksi

Step 4: Calculate the Pall of the HP 14 × 89

Pall = 11.91 ksi × 26.1 in2 = 310.9 k

Normal stress on steel would allow Fbs = 27,000 psi; however, because of buck-

ling, the allowable stress was lowered to 11.91 ksi.

2.6.1 Combined Stress

Combined stress occurs when a material section has a condition causing normal and

bending stress at the same time. One example would be a pipe full of water and

attached to the side of a post or column. The pipe and water weight (P) create twostresses: axial (normal) stress from the weight of the pipe over the cross-sectional area

of the post and a bending stress from the weight of the pipe (P/A), and e eccentricity


distance away from the centerline of the post creating a bending moment and stress

on the section modulus of the post Pe∕S. See Figure 2.16.Combined stress = P∕A + Pe∕S, units in psi or ksiUsing this same scenario, given the information to follow, determine the combined

stress in psi.

Given: 4 × 4 S4S wood posts at 12 ft on center

4′′ STD pipe full of water

Assume the pipe is adequately attached to the side of the post

Step 1: Determine the weight of the pipe and the water inside the pipe. The dimen-

sions and properties of 4′′ standard pipe are found in Appendix 2. It has anID = 4.026′′ and an OD = 4.50′′. The inner diameter (ID) will be used for

the water volume and the outer diameter (OD) will be used for the distance

e eccentricity. The pipe itself weighs 10.79 plf.

Water weight: Either convert the unit weight of water to cubic inches

or convert the inside pipe volume to cubic feet. Let’s use the former

method. Water weighs 62.4 pcf, therefore, 62.4 pcf∕1728 in3∕ft3 =0.03611 lb∕in3 of water.

The inside volume of the pipe is the cross-sectional area × 12′′ lengthof pipe (1 ft).

𝜋(4.026′′)2∕4 = 12.73 in2 × 12 in = 152.76 in3

Water weight in a 12′′ piece of pipe = 152.76 in3 × 0.03611 lb∕in3 =5.51 plf.

Add the weight of the pipe per l linear foot, 5.51 plf + 10.79 plf =16.31 plf.

Step 2: Calculate the weight of 12 ft of pipe

16.31 plf × 12 ft = 195.7 lb

Step 3: Calculate the distance from the center of the pipe to the center of the post.

e = 1

2(pipe OD) + 1

2(post width), e = 1

2(4.50′′) + 1

2(3.50′′) = 4.00′′

Step 4: Calculate the axial stress due to the weight of the pipe on one post loading

the cross-sectional area of the 4 × 4 post.

PA

= 195.7 lb

3.5′′ × 3.5′′= 15.98 psi

Step 5: Calculate the bending stress on the post due to the weight of the pipe edistance from the post causing a bending stress. The section modulus of

the 4 × 4 post is 7.146 in3

PeS

= 195.7 lb × 4.0′′

7.146 in3= 109.54 psi

2.6 BEAM DESIGN 43

FIGURE 2.17 Combined stress.

Step 6: Combine the two stresses similar to Figure 2.16. Figure 2.17 depicts this

combined stress:

Combined stress = 15.98 psi + 109.54 = 125.52 psi

This does not consider how the pipe is attached to the post. That would have to be

designed separately.

Combined stress can also occur when a column or strut that are supporting an

axial load encounter a concentrated or uniform lateral load against the weak or strong

axis. For instance, when a steel beam is used as a strut in a cofferdam, its original

purpose is to resist the axial force from the soil loading that is pushing from each side.

However, during the excavation, what would happen if a cubic yard of soil weighing

approximately 3800 lb fell on the beam. Not even considering the dynamic force of

the 3800 lb force, the strut would have to support the axial force at the same time the

lateral force of the dropped soil applied a bending moment to the beam. The sketch

in Figure 2.18 is an FBD of how these two forces would affect the beam.

Similar to the example above, the axial stress must be added to the bending stress.

The difference in this example is that the bending stress moment is represented as

P1L∕4. The axial force is still represented by P/A. Another way to explain the com-

bined stress analysis in a case like this would be to analyze each stress as it relates to

its allowable stress. In other words, the axial force is a buckling stress case and the

bending moment is a bending stress case. Therefore, the allowable buckling stress

needs to be known and the allowable bending stress should be determined (between

0.6Fy − 0.67Fy).

P∕A∕ 𝜋2E

(kl∕r)2 × FOS+

PL∕4∕S0.60 × 36,000 psi

< 1.0

The theory behind this method is that if the two individual stresses are within their

own ranges, they will pass the stress test; or, if together, they do not exceed 100% of

the combined allowable stresses, they should be within the limits.

FIGURE 2.18 Combined stress on a beam.


Example 2.9 Determine the combined stress of a 6′′ standard pipe strut. The strut

is 15 ft long, the axial force from the soil is 30 k and there is a cable in the center of

the pipe holding a 1500-lb trash pump. The FBD shown in Figure 2.18 depicts this

situation.

Since the strut is 15 ft long, the cable is connected 7.5 ft from each end. The prop-

erties needed for the pipe are

A = 5.58 in2

S = 8.5 in3

r = 2.25 in

Determine the axial stress compared to its allowable buckling stress and then

determine the bending stress compared to the allowable bending stress on A36 steel

assuming 0.6Fy:

30,000#5.58 in2

+

1500# (15′′)4

8.5 in3≤ 1.0

Π2 29,000,000 psi[(15′ × 12

)2.25

]2

× 1.92

= 23.293 psi

(0.60 × 36,000 psi) = 21,600 psi

5376 psi

23,293 psi+

7941 psi

21,600 psi= 0.60 ≤ 1.0 (OK)

In this example, the pipe would not fail in combined stress because, when com-

bined, the axial stress and the bending stress do not exceed more than 100% of the

combined allowable stresses.

CHAPTER 3

TYPES OF LOADS ON TEMPORARYSTRUCTURES

3.1 SUPPORTS AND CONNECTIONS ON

TEMPORARY STRUCTURES

There are a few luxuries we have as construction managers in temporary structuresdesign. The first one is that the construction manager and designer have a great dealof flexibility in system selection during the design phase. There are typically manydifferent solutions to solve the same problem. These solutions are generally comparedto one another and costs are placed on each. The best solution usually is the one thatcosts the least, is most schedule efficient, and does not come with any safety andhealth risks to the workers, the general public, or property.

Another luxury we have as construction managers is the way we support theindividual members of our temporary structures. In statics, it is very important howthe ends of beams or columns are supported when determining the amount andtypes of stresses incurred from bending moments, shear, buckling, and deflection.In temporary structures, supports are almost always considered to be simplysupported or hinged (Figure 3.1). This means that when a beam rests on its support(another beam), they are not welded, bolted, or otherwise joined/fixed together. Onemember is simply resting on another, and, when the loads are applied, the beam isfree to rotate about the point where they make contact. When supports are boltedor welded (fixed connections at the ends), the beam’s maximum moment can bereduced from wl2∕8 to wl2∕12, but the labor involved with making the connectionbecomes a large contributor to the overall costs on the temporary structure.

When beams are simply supported, their deflection pattern begins and ends abovethe supports. When this is the case, we have to use the more conservative moment anddeflection formulas. They are more conservative because they produce more moment

45

46 TYPES OF LOADS ON TEMPORARY STRUCTURES

FIGURE 3.1 End connections.

and more deflection. Two examples of these formulas are

wl2

8

5wl4

384EI

These two formulas will produce higher moments in the beam and higher deflec-tion values than if the beam is supported by multiple supports or fixed at either orboth ends. The following formulas are used to calculate maximum bending momentand maximum deflection in beams with three or more spans:

wl2

10

wl4

1740EI

Columns are somewhat similar. The distance needs to be analyzed between wherethey make contact with the ground and what they are supporting. These points ofcontact are said to be hinged at the top and bottom so that when the axial force isapplied to the column and the column’s reaction is to buckle, it does so at the pointsof contact. In this case the effective length is equal to the full length of the column.In this text, wewill assume hinged connections for all contact points and support loca-tions. In a later chapter, connections will be discussed briefly so the student can gainsome understanding of how other types of connections affect temporary structuresdesigns. Even though hinged connections are the focus of this textbook, the studentshould understand the different types of connections that affect stress in columns.In mechanics, the student learns the following end conditions and applies the appli-cable k value. As mentioned in Chapter 2, k is the multiplier that is used, dependingon the connection.

Both ends hinged, k = 1.0.

One end hinged, one end fixed, k = 0.70.

Both ends fixed, k = 0.50.

Figure 3.2 illustrates what happens to le when lateral braces are added to a post.P2 would be used if there were only a top and bottom support. However, if an

3.1 SUPPORTS AND CONNECTIONS ON TEMPORARY STRUCTURES 47

FIGURE 3.2 Sample post supports.

additional support was added in the center of the post, P1 would represent the new

effective length.As a simple example, if a column is 12 ft tall, resting on a concrete slab and sup-

porting the underside of a beam, the effective length (le) is equal to 12′ × 1.0 = 12′.If we made a connection between the supporting slab and the post that was adequate

for fixity, the effective length (le) would then be 12′ × 0.70 = 8.4′.Another consideration in supporting temporary structures is not only what type of

support but how many supports are used under a continuous single member. In statics

and strength of materials, this is the difference between simple moment calculationsand an indeterminate structure that has more than two unknown reactions (supports).

When a member spans over more than two supports, much time can be spent deter-

mining the reaction values. In this book, these two types of support systems are cate-gorized into either a simply supported beam or a continuous beam over two or more

supports. This concept will be discussed in more detail when beams, columns, and

struts are analyzed in other chapters.

3.1.1 Forces and Loads on Temporary Structures

Axial Axial forces are common in columns, struts, and tension members such as

coil rods and anchors. In a simple case, the force is directed concentrically throughthe center of the supporting member. Other cases have the loads off center to the

centerline of the axial member. In this case, combined stress should be considered.

Combined Loads Combined stress is when there is an eccentric load causing

axial (normal) stress and bending stress simultaneously. The two stresses are com-

bined (added or subtracted) to result in a total stress at the extreme fiber. In temporarystructures, the largest combined stress is the most important, as this is the worst-case

situation. This concept was discussed in length in Chapter 2.

Compression Compression has several considerations when it comes to wood

members. Forces can be applied parallel to grain or perpendicular to grain as well as

induce buckling stress. The first two stresses are basicallyP/A-type stresses. Bucklingstress, however, is a reduced value to normal stress, which takes into account the

length of the column or strut and its radius of gyration or least dimension in cross


FIGURE 3.3 Compression zone.

section depending whether steel or wood is used. Figure 3.3 illustrates the area of thecompression zone.

Compression in steel is treated the same as wood; however, the concern is not asrelevant. Allowable stress on A36 steel can be as much as 75–90% of Fy, its yield,or 27–32.4 ksi.

Tension Tension forces can be produced by hanging supports in coil andDWIDAG-type rods or tieback, soil, and rock anchors using rods or high-strengthstrand.

Concentrated and Uniform Loads Concentrated loads occur when one beamelement’s load path travels through another beam element. The bottom of one beam’sflange applies load to the top of another beam’s flange. If the beam spacing is a quarterof the span or more, this load type can be considered a uniform load. If the spacingis less, one can consider this multiple concentrated loads.

Uniform and Varying Loads Uniform loads are probably the most commonforces in temporary structures because they represent a continuous force that actsperpendicular to the supporting face. These forces come from water, concrete, wind,and soil. Some of these uniform forces are constant and some are varying.

3.1.2 Materials—How Different Materials Create Different Forces

Water Water can produce both uniform and varying forces. If the water pressureis static on a cofferdam, for example, then it is considered varying, with the highestforce being at the bottom and reducing to zero at the top. This condition is shown inFigure 3.4. Water pressure, in this case, is the product of the depth and the unit weightof water.

Static Water Pressure Static water projects a triangular pressure diagram andacts perpendicular to the supporting face. The pressure is a direct result of the depthof water and the unit weight of water (always 62.4 pcf in this book). The maximumpressure (Pmax) for a 25-ft deep (h) body of water in its static state is

Pmax = 𝛾waterh

Pmax = 62.4 pcf × 25 ft = 1560 psf


FIGURE 3.4 Water pressure diagram.

Flowing Water Flowing water will apply a uniform force of the side of thestructure that is obstructing flow, such as a cofferdam or permanent pier or column.Figure 3.5 shows how high water levels can get in construction. The river shown inthis figure reached its peak flood elevation just 4 months before construction of itsreplacement bridge began. This would be a 3-year project.

Figure 3.6 provides a simple water pressure diagram. To calculate flowing waterforce, the following formula is used from physics:

P = cd𝛾[(√2

)∕(2g)

](units in psf)

where P = pressure in psf

Cd = drag force coefficient of the obstruction

𝛾 = unit weight of water (pcf)√= velocity of flowing water (ft/s)

g = acceleration of gravity at Earth’s atmosphere = 32.2 ft∕s2

The drag force coefficient, Cd, takes into account the friction created by the wateragainst the side of the obstruction as the water passes. The drag force will vary based

FIGURE 3.5 Flooding waters on a bridge pier.


FIGURE 3.6 Flowing water pressure diagram.

on the shape of the obstruction. The drag force can range from 1.2 for cylinders such

as steel pipe to 2.0 for Z-shaped sheet piling with convolutions in the sheets created

by the Z shape.

Example 3.1 Determine the pressure from a flowing river with the following char-

acteristics. Then determine the resultant if the water is 20 ft deep.

Velocity of water = 12 ft∕sUnit weight of water = 62.4 pcf

Drag coefficient of 1.3

Step 1: Calculate the pressure in psf.

P = 1.3(62.4 pcf)

[122

2(32.2 ft∕s2

)]

P = 181 psf

Step 2: Since the force is a rectangular load, the resultant is in the middle or the

flowing water (see Figure 3.7). The resultant needs to be calculated by

multiplying the pressure times the depth of the water:

R = P × Depth R = 181 psf × 20 ft = 3620 plf of flowing water

Remember, the drag coefficient can vary from 1.2 to 2.0.

FIGURE 3.7 Resultant force location.


FIGURE 3.8 Triangular force on an incline.

Loads on Inclining/Sloped Surfaces Pressure from water is developed from

the unit weight of water and the water depth, and the direction of the pressure acts per-

pendicular to the surface of the obstruction or dam. What happens when the surface

of the obstruction or dam is sloped instead of vertical? Does the pressure increase?

No, because we already determined that the pressure magnitude comes from the unit

weight and depth of water. Does anything else change? Yes, the length of supporting

surface becomes the sloped surface of the triangle. What does this mean?

The static water example above used a vertical surface as an obstruction in order

to calculate the water pressure of 1560 psf. The resultant of this pressure over the

25-ft-tall dam is determined by calculating the area of the triangular pressure diagram

(see Figure 3.8).

The resultant is 1560 psf × 25 ft∕2 = 19,500 lb acting perpendicular against the

surface one third of the distance from the bottom.

If the surface was sloped at a 45∘ angle, still maintaining the 25-ft water depth,

the pressure would remain the same at the bottom. However, the pressure triangle

would lengthen, thus increasing the resultant magnitude. This occurs because when

the resultant is recalculated, the length of the pressure triangle is now 25 ft × 1.414

(the square root of 2) = 35.35 ft.

R =1560 psf × 35.35 ft

2= 27,573 lb

The resultant increased by a factor of 1.4, which happens to be the 45∘ angle squareroot of 2.

Concrete Concrete is a unique force in that it is liquid for a period of time but

then begins to set and support itself eventually. Concrete can be supported with a

vertical form in which the force acts laterally against the form face. This force is

called hydrostatic pressure because it is present for a period of time until the concrete

begins to set and support itself. This period of time is between 2 and 6 h, depending

on mix design and admixtures used to retard or slow down the initial setting of the

concrete.

Concrete can also be supported as an elevated slab element where its force is a

gravity load acting downward to the ground. Both of these cases are discussed below.

Concrete supported as a gravity force—Concrete weighs approximately 150 pcf.

When the concrete is reinforced, 5–10 lb can be added to the concrete weight. In this

text, reinforced concrete weighs 160 pcf (150 + 10 = 160 pcf). The thickness of theconcrete element is very important because in order to convert the load to psf, the unit


FIGURE 3.9 Reinforced concrete column.

weight must be multiplied to the thickness in feet (see Figure 3.9). Let’s consider

Example 3.2.

Example 3.2 Determine the maximum load in psf of 16-in-thick reinforced

concrete.

P = 160 pcf × 16′′∕12′′∕ft = 213.3 psf. The force is now in PSF after multiplying

the unit weight in pcf times the thickness in feet. The sketch in Figure 3.9 depicts a

12′′ × 12′′ × 16′′ thick slab section.

In Chapter 10, a live load will be added to the dead load in order to come up with

the actual load on the forms. The dead load takes into account the concrete and rebar

weight and the live load is the estimated weight of the tools, equipment, and workers

(see Figure 3.10).

Concrete as a hydrostatic force is probably the most common load in construction.

If concrete was treated as a material that is always liquid, like water, the pressure

would be 150 pcf × h. However, concrete does not stay liquid for a long period of

FIGURE 3.10 Concrete placement on a bridge superstructure.


time; it begins to set and support itself after a period of time (typically 2–6 h). For this

reason, special formulas are used to determine the maximum pressure. The pressure

diagram for concrete is either trapezoidal or triangular. The trapezoidal load is more

common. The triangular pressure diagram is used when the concrete is considered

fully liquid. Both will be discussed in Chapter 8.

The following pressure diagrams in Figure 3.11 show what trapezoidal and trian-

gular loads of concrete hydrostatic pressure would look like.

Pmax

Hydrostatic (concrete)

Pmax

h

Full Fluid Pressure

h

FIGURE 3.11 Hydrostatic and full liquid pressure.

Wind Wind projects a horizontal uniform load from top to bottom until it reaches a

certain height. This is indicative of a wind force on a column form or wall panel.Wind

loads are calculated in pounds per square foot (psf) but can be converted from miles

per hour (mph) of wind speed in a particular geographical zone. Table 3.1 shows how

pressure (psf) corresponds to different wind speeds in mph. The table also relates

this force to the force generated against a 4 × 8 sheet of plywood and against an

TABLE 3.1 Wind Speed Converted to Wind Pressure

mph psf

Force on 4 × 8 Sheet

of Plywood (lb)

Force on Average

Size Person (lb)

10 0.27 8.6 2.2

20 1.08 34.6 8.6

30 2.43 78 19.4

40 4.32 138 35

50 6.8 216 54

60 9.7 311 78

70 13.2 423 106

80 17.3 553 138

90 21.9 700 175

100 27 864 216

110 33 1045 261

120 39 1244 311

130 46 1460 365

140 53 1693 423

150 61 1944 486


FIGURE 3.12 Wind pressure diagram.

TABLE 3.2 Wind Pressure at Different Height Zones

Height Zone (height above grade) (ft) Wind Pressure (psf)

0–30 20

30–50 25

50–100 30

Over 100 35

average size human being’s body. If someone has ever experienced lifting up a piece

of plywood in the wind, he would understand the force being described.

When a project is constructed in a particular region, the designer and contractor

can obtain wind force records common to the area. The designer of the permanent

structure needs this information for the permanent design. Likewise, the contractor

and his designer need this information for any temporary structure design.Wind force

can affect forms for walls and columns, rebar for walls and columns, and any other

structure above the ground that would be affected by wind forces.

State Department of Transportation and the AISC have developed best practices

for column rebar guying. This will be covered in Chapter 12. However, the under-

standing of the wind forces created on a rebar cage that stands over 20 ft in height is

very important. The best practices typically increase the wind speed (pressure) as the

column gets higher. For instance, a 50-ft–high column designed for a 90-mph wind

speed would be designed for 21.9 psf in the first 30 ft. The remaining 20 ft would then

see a higher pressure. The increase in the upper sectionmay be between 5 and 10mph,

thus 36–31 psf. The pressure diagram in this case would look like the Figure 3.12.

Table 3.2 is an example of wind forces in California rounded to the nearest 5 PSF.

When the two charts are compared, it is obvious how the wind force used on these

projects is the maximum wind speed that is recorded in the region over a long period

of time. Even though reaching top speeds may not be likely, the designer must antic-

ipate the worst-case scenario. As one can see, Table 3.2 uses wind pressure forces

between 20 and 35 psf. Looking at Table 3.1, this is the equivalent of approximately

80–120 mph winds.

Example wind force will be illustrated here. Using the chart in Table 3.2, a wind

load will be applied to a vertical wall.

Example 3.3 Calculate the resultant force of a 50-mph wind on a 20-ft-high ×15-ft-wide wall. Draw a pressure diagram like that depicted in Figure 3.13.


FIGURE 3.13 Wind pressure diagram.

Step 1: Determine the force from the wind as a 12′′ strip load as the resultant:

R (12′′ strip) = 6.8 psf × 20 ft = 136per linear foot of wall

The resultant force of this 20-ft strip load is 10 ft from the bottom.

Step 2: Determine the resultant force of the 15-ft-wide wall.

R(wall) = 136 plf × 15 ft = 2040 lb located 10 ft from the bottom

and 7.5 ft from the left and right. Draw a resultant diagram like that in

Figure 3.14.

FIGURE 3.14 Resultant diagram.

Soil Pressure Soil pressure places a varying load laterally on a support of exca-

vation (SOE). The forces from soil pressure have many variables, but the resultingpressure can be estimated using several different theories on soil pressure. This bookwill use the most common theories of soil mechanics to illustrate the pressure applied

to an SOE from different types of soil. The theories of Coulomb and Rankine, whichuse the results ofMohr’s circle, will be simplified to the extent that a great deal of timeand effort are not spent explaining the origins of the theories. Like many examples

in this book, the concepts can be understood without all the details that the originalauthors used in their calculations. This subject will be studied further in Chapter 5.

Other effects on shoring, besides the soil pressure, are the water table and equipmentsurcharge loads.


FIGURE 3.15 Soldier beam and lagging system.

The water table elevation can affect one of four things:

Lighten the effective weight of a sandy, granular soil.

Increase the unit weight of a clayey soil.

Introduce water into the excavation through boiling or piping upward through thesubgrade.

Add additional water load to the shoring design.

Most excavations have construction equipment working near or against the top ofthe shoring system. The weight of the equipment, as a gravity force, projects a lateralforce on the shoring system. This force is added to the other lateral forces from thesoil to achieve the maximum pressure the system will experience. Figure 3.15 showsa basic soldier beam and lagging support system. This type of systemwill be designedin Chapter 6.

Soil Supported as a Hydrostatic Force Like concrete, soil that is supportedlaterally projects a horizontal force that is triangular, rectangular, and sometimestrapezoidal. The soil properties play a large role in the amount of pressure appliedat the bottom of the pressure diagram. Later in this book, soil pressure formulas willbe studied. For the purpose of this chapter, the understanding of the different pressurediagram shapes is the most important. Depending on the theory used, a soil pressurediagram can be triangular, rectangular, or trapezoidal. Some soil pressure examplesare shown in Figure 3.16. These diagrams range from rectangular forces to triangu-lar and forces in between. Soil pressure is probably the most difficult to representof all temporary structure forces. Chapter 5 will explain the basics of the theoriescombined.


FIGURE 3.16 Different soil pressure diagrams.

Soil as a Vertical Force Soil supported is a gravity force since soil weighs

approximately 90–140 pcf. The thickness of the soil is very important because in

order to convert the load to psf, the unit weight must be multiplied by the thickness

in feet. For example, let’s consider the following soil load.

Example 3.4 Determine themaximum load in psf of a 6-ft-thick soil columnweigh-

ing 110 pcf.

110 pcf × 6 ft = 660 psf. The force is now in psf after multiplying the unit weight

in pcf times the thickness in feet. Figure 3.17 depicts a 12′′ × 12′′ × 6 ft soil column.

FIGURE 3.17 Soil as a column.

Supporting Existing Structures Existing structures need to be supported dur-

ing retrofit and expansion projects. It is very common to have to support bridge

components during the construction of seismic retrofitting. Sometimeswhole sections

of permanent support systems such as piers and columns have to be held in place dur-

ing the removal and reconstruction of the new supports. After the 1989 earthquake

in the San Francisco Bay Area, billions of dollars were spent reinforcing, removing,


FIGURE 3.18 Temporary support of an existing structure.

and replacing bridge elements in order to withstand a reoccurring earthquake of sim-

ilar or greater magnitude. Figure 3.18 shows a structure being supported at a water

treatment plant project that is being upgraded and increased in treatment capacity.

Oftentimes, buildings or water and wastewater facilities have to be expanded for

increased use beyond their original design use. Buildings need to handle more people,

or a water facility needs to treat and pumpmore gallons per day. If a new structure has

to be added next to an existing structure, it may be necessary to support the existing

structure from the risk of being undermined. Also, elevated slabs, beams, or columns

may have to be supported in order to lengthen or increase their cross-sectional areas.

When existing structures are involved, the unit weight of the materials being sup-

ported should be used. If the material type is questionable, use the more conservative

approach. When there is a combination of different materials, the unit weight of the

heaviest material should be used.

CHAPTER 4

SCAFFOLDING DESIGN

Scaffolding in construction is a common method of access to elevated areas of work.Scaffolding has been used for centuries in forms of earthen fills, timber, and bambooand, in the most current century, steel and aluminum combined with wood members.Since scaffolding is so regulated in the United States and other progressive coun-tries, the design of its components are typically standardized and performed onlyby the manufacturers of the systems themselves. When contractors are in need of ascaffolding system, they typically buy or rent preengineered systems.

This chapter does not imply that a contractor would have an engineer designcustom scaffolding. Scaffolding is simply used in this book to illustrate simple engi-neering practices, load paths, and factors of safety.

4.1 REGULATORY

The Occupational Safety and Health Administration (OSHA) has the jurisdiction andsets the standards for scaffolding in the United States. Since scaffolding supportshuman beings in the workplace, higher standards are set to protect the health andsafety of the workers and the general public. This chapter will discuss the safetyfactors that are required in scaffolding, cover a complete design of a scaffold, anddiscuss the different types of loads for which scaffolds are rated. The reader shouldconsult OSHA documents for specifics on correct installation of a scaffold system.

4.2 TYPES OF SCAFFOLDING

The two types of scaffolding covered in this chapter are (1) tube and coupler and(2) premanufactured frames. The tube-and-coupler type will be used to teach scaffold

59

60 SCAFFOLDING DESIGN

FIGURE 4.1 Tube and coupler.

design and the premanufactured frames will be discussed because they are the most

popular type of scaffolding used today.

Tube-and-couple scaffolding is made up of individual pieces of pipes and couplers

as shown in Figure 4.1. Wood or aluminum scaffold planks are added to complete

the system. This scaffolding is good for irregular shapes of buildings and structures

because each piece is selected and all are put together tomaximize efficiency. Because

we can design each component as we follow the load path, and the system is custom

assembled, tube-and-couple scaffolding provides a perfect design example.

Premanufactured frames are the most common type of scaffold because the com-

ponents are a fixed dimension and predesigned. As long as the pieces are assembled

per the manufacturer’s instructions and loaded within the load ranges, the scaffold is

safe and follows OSHA regulations. Figures 4.2 and 4.3 illustrate some of the com-

ponents of premanufactured frames.

The assembly shown in Figure 4.3 is connected to similar assemblies to make full

systems as shown in Figure 4.4. These assemblies can be integrated into multistory

systems for all types of construction.

FIGURE 4.2 Various frame types.

4.3 LOADING ON SCAFFOLDING 61

FIGURE 4.3 Assembly of frames and cross braces.

FIGURE 4.4 Scaffolding used for building construction.

4.3 LOADING ON SCAFFOLDING

Scaffold loading can be determined by a duty schedule based on type of work and use

of scaffold, or it can be determined by the actual weights for which the scaffold will be

used. The duty schedule is a scaffold rating that determines a design load in psf based

on the type of construction for which the scaffold will be used. Table 4.1 describes


TABLE 4.1 Scaffold Duty Schedule of Loads

Duty Level Design Load (psf) Type of Work

Light duty 25 Painting, cleaning

Medium duty 50 Carpentry, light framing

Heavy duty 75 Concrete, steel, masonry, demolition

three different levels of scaffold use. The types of use dictate the typical personnel,materials, and tools that are used by the trades that perform the prescribed type ofwork. This book will use this information to determine the scaffold loading.

4.4 SCAFFOLDING FACTORS OF SAFETY

As mentioned above, scaffolding tends to require a larger factor of safety (FOS) thanmost other temporary structures. OSHA oversees the requirements for scaffold andimposes more stringent conditions than the standard 1.5 : 1 to 3 : 1 FOS. For scaffoldin this chapter, the following FOS will be used for design. The temporary structuredesigner should consult specific local requirements before proceeding with scaffolddesign.

Scaffold structural steel and lumber members 4.0

Wire rope 4.0–6.0

4.5 SCAFFOLD COMPONENTS

4.5.1 Planking

Planking is the part of the scaffold with which the workers, materials, and tools makedirect contact. The planking spans from bearer to bearer and transfers the loads fromthe type of use to the bearers. The two most common materials for planking are woodand aluminum. This book will use wood planking in its design examples. Aluminumplanking is premanufactured and charts are available through themanufacturer, whichthat prescribe the usage limits and maximum allowable spans.

4.5.2 Bearers (Lateral Supports)

Bearers, typically pieces of pipe, support the planking and transfer the planking loadto the posts with special connectors. The width of the bearers determines how muchwork space is available. The more work space, the more efficient the workers will be,but the more expensive is the scaffold in material and labor costs, both to install andto remove. The shorter the bearers, the lower the scaffold cost, but with less workspace, which translates to lower worker efficiency.

4.5.3 Runners

Runners span from post/bearer to post/bearer. They traditionally do not carry anyloads except to transfer lateral stability from post set to post set. In the design stage,the runners will not be designed for size.

4.6 SCAFFOLD DESIGN 63

FIGURE 4.5 Typical scaffold section for design examples.

4.5.4 Posts

Posts transfer the load from the bearers and take all the concentrated loads fromall levels to the ground or supporting surface. Post length is determined by howmuch room the contractor needs between the levels of the scaffold. This distancecan also be dictated by the distance between the floors of the building so that thereis a scaffold level at each floor level. For simplicity, posts are made up of the samepipe material as the bearers and runners. Figure 4.5 shows a typical scaffold sectionof seven levels. This will be the basis for our design example for tube-and-couplerscaffold.

4.5.5 OSHA

As mentioned earlier, this book is not designed to cover the rules and regulations setforth by both state and federal OSHA. The intent is strictly a design aid. For moreinformation on OSHA rules and regulations, go to www.osha-safety-training.net.

4.6 SCAFFOLD DESIGN

Scaffold can be designed with twomethods. The first method is when we know all thespacing of the different parts and we know the material type. From this information,the material size is determined. The secondmethod is when thematerial size is knownand the spacing of each component must be determined. In this book the first methodmentioned will be used.

http://www.osha-safety-training.net


Example 4.1 Scaffold Design with Tube and CouplerDesign a scaffold with A36 standard steel pipe to carry the load of a medium-duty

construction operation with the following features and material specifications:

Medium-duty loading = 50 psf

Three-level scaffold

Scaffold width front to back = 4 ft

Distance between posts along the front and back = 7 ft

Distance between levels = 9 ft

Planking: Douglas Fir Construction Grade, Fb = 900 psi (includes a 4 : 1 FOS)

Steel pipe A36 standard schedule 40, Fb = 9000 psi (includes a 4 : 1 FOS)

Step 1: Plank Design—Since medium duty is the required loading, 50 psf will be

loaded on the planking to begin the design. When planking is designed, a 1-ft width

is selected in order to keep the units from psf to plf. In other words, multiply the Pmax

by 1 ft so w units are in plf.

Pmax = 50 psf

w = 50 psf × 1 ft = 50 plf

Draw an FBD of one span of planking from bearer to bearer, as shown in

Figure 4.6.

FIGURE 4.6 Simple planking FBD.

The length of planking is equal to the distance between the bearers or the distance

between the post sets along the front or back. The given information above allows for

a 7-ft planking span. Since plank design is based on 1-ft increments, w = 50 plf.

Determine theminimum thickness of the planking by combining themost common

bending stress formulas. These are

M = wL2

8

f = MS

S = bd2

6

Since w and L are known, moment can be calculated by using the following

formula:

M = (50)(7′)28

= 306.3 ft-lb


FIGURE 4.7 Cross section of a 12′′ plank section.

Now we know that Fb is 900 psi based on the given information regarding the

material. Also, since a 12-in (1-ft) strip of scaffold is the basis of design, shown

in Figure 4.7, then b in the section modulus formula is set at 12 in. Therefore, the

bending stress formula can be completed with only one unknown, d.Very Important: In order to keep the units correct, the max moment must be mul-

tiplied by 12′′∕ft in order to work with the unit of inches in section modulus. It will

look like this:

900 psi =(306.3 ft-lb)(12′′∕ft)

(12′′)(d)2∕6

solve for d, where d = 1.43 in.

Conclusion: The plank thickness must be a minimum of 1.43′′ thick, which wouldbest be accommodated using a 2X nominal plank, which measures an actual 1.5′′ inthickness, slightly greater than the required 1.43′′.

Step 2: Determine the minimum size bearer using A36 standard steel pipe.

Draw an FBD of one bearer spanning from a front post to a back post. Figure 4.8

shows a sample FBD of this condition. Calculate the uniform load on the single bearer

by multiplying the design load by the plank span. w = 50 psf × 7 ft = 350 plf.

FIGURE 4.8 FBD of a single bearer.

Using the same bending stress formulas, determine the minimum standard steel

pipe size for a bearer. First calculate the allowable bending stress of A36 steel with

a 4 : 1 FOS. Since the steel pipe chart in Appendix 2 gives the section modulus for

each size, calculate the minimum S for the pipe.

Fb =fy4.0

36 ksi

4.0= 9 ksi (9000 psi) = Fb

M = wL2

8

f = MS

M = (350)(4ft)2∕8 = 700ft-lb, therefore, 9000psi = (700)(12′′∕ft)∕S, Smin =0.933 in3

Select the standard steel pipe with a section modulus (S) greater than or equal to

0.933 in3.

From the pipe chart in Appendix 2, the best pipe would be the standard 21

2-in pipe.


FIGURE 4.9 Combined stress in pipe posts.

Step 3: Determine the minimum size steel post to carry three levels of load while

still staying within the combined stress limits. There are two modes of failure for

a post, buckling failure and combined stress failure. Buckling can occur when the

combined load of all the scaffold levels exceeds the buckling resistance of the pipe.

The buckling resistance of a pipe is dependent on a higher radius of gyration (r)combined with a low unsupported length (L). Combined stress occurs when the axial

stress on the pipe (P1∕A) is added to the effects of the bending stress caused by the

eccentric location of the bearer as it is attached to the post (P2e∕S). This condition isshown in the Figure 4.9. Buckling typically fails before combined stress, so the post

should be designed for buckling and then checked for combined stress.

The first task is to determine the load on the post for both conditions. P1 is the

accumulation of all the levels to the lowest post on the ground. P2 is the load of one

level to the bearer and onto the post. These loads are calculated as follows:

P1 = 3 levels × plank length + 1

2bearer length × Pmax

P1 = (3)(7′)(2′)(50 psf) = 2100 lb

P2 = P1∕3 levels or plank length × 1

2bearer length × Pmax

P2 = 2100 lb∕3 = 700 lb or (7′)(2′)(50 psf) = 700 lb

Buckling Stress for Steel Sections

Chapter 2 briefly described buckling in columns. In order to size the post based on

buckling stress, the designer should select a standard piece of pipe from the chart for

a trial-and-error process. Once the pipe is selected, record the values for r, A, and S(radius of gyration, cross-sectional area, and section modulus, respectively).

The pipe selected in this example will be an 11

2standard steel pipe.

r = 0.623 in

A = 0.799 in2

S = 0.326 in3

Fbs =𝜋2E

(kL∕r)2 × FOS

Fbs =Pall

A


where Fbs = allowable buckling stress in psi

E = modulus of elasticity of the steel, which is 29,000,000 psi unless

noted otherwise (psi)

K = connection of the pipe to the bearer, which will always be a hinged

connection (1.0 multiplier)

L = unsupported length of the pipe from the ground to level 1 (in)

r = radius of gyration of the pipe selected (in)

FOS = 4.0 factor of safety for scaffolding

𝜋 = 3.1416

Fbs =𝜋2(29,000,000 psi)

[(1.0)(9′ × 12′′∕ft)∕0.623′′)]2 4.0Fbs = 2381 psi allowable buckling stress

This means that instead of an allowable stress of Fy∕4.0, 36,000 psi∕4.0 =9000 psi, due to buckling only 2381 psi is allowed.

Now set Fbs = Pall∕A, and solve for P1 allowable.

2381 psi = Pall∕0.799 in2, Pall = 2381 psi × 0.799 in2, Pall = 1902 lb

The actual load that was calculated earlier was 2100 lb. Since 1902 lb is less than

2100 lb, this pipe DOES NOT WORK.If the two values were way off, then the next pipe selected would maybe be a

couple sizes larger. However, since they were only 200 lb off, the 2′′ standard pipe

can be the next selected for the trial-and-error test.

2′′ standard pipe:

r = 0.787 in

A = 1.07 in2

S = 0.561 in3

Recalculate buckling stress; the only change is to radius of gyration; the length

and other variables and constants are still the same.

Fbs =𝜋2(29,000,000 psi)

[(1.0)(9′ × 12′′∕ft)∕0.787′′ )]2 4.0Fbs = 3800 psi allowable buckling stress

3800 psi = Pall∕0.799 in2, Pall = 3800 psi × 1.07 in2, Pall = 4066 lb

4066 lb > 2100 lb (OK)

The 2′′ standard pipe is OK for buckling; now let’s check this same pipe for

combined stress.

Combined Stress: As mentioned earlier, combined stress is the combination of

axial stress and bending stress at the same time. These two stresses act on the same


pipe section, just below the connection of the bearer to the pipe post. In this example

there are three levels of scaffold bearing down on the lowest level post and, at the same

time, the first level load is bending the same pipe because the bearer is eccentrically

loading the post at the same point. Combined stress is not calculated until the pipe

has been selected for buckling. Combined stress is calculated as follows:

FCS =P1

A+P2e

S

This answer is in psi because normal stress and bending stress, both in psi, are

being added together:

where P1 = combined loads of all levels on a single lower post

A = cross-sectional area of the post

P2 = load from a single level on the bearer that is eccentrically loading the

lower post

E = modulus of elasticity of the steel pipe, typically 29,000,000 psi in this

text

S = section modulus of the post

For buckling, the load for all three levels was needed. This value is still going to

be used for P1. P2 is necessary for the eccentricity load on the post caused by the

bearer.

P2 = 700 lb

Finally, the eccentricity distance is needed in order to calculate the bending stress.

e = 1

2(post OD) + 1

2(bearer OD)

Post (2′′ standard) OD = 2.375′′

Bearer (21

2

′′ standard) = 2.875′′

e = 1

2(2.375′′) + 1

2(2.875) = 2.625′′

Using the section properties for the standard 2′′ pipe from above and P1, P2, and e,calculate the combined stress on the post:

Fcs =(2100 lb

1.07 in2

)+[(700 lb) (2.625′′)

0.561

]1963 psi + 3275 psi = 5238 psi

Allowable bending stress for any scaffold component is Fy∕4.0 FOS =36,000 psi∕4 = 9000 psi.

Therefore, Fcs must be below 9000 psi; 6486 psi < 9000 psi (OK)

Conclusion to scaffold design:

Planking: minimum 11

2′′ thick Douglas fin (DF) construction grade

Bearer: 21

2′′ standard steel pipe

Posts: 2′′ standard steel pipe


It is the designer’s option whether to size the posts differently for the second and

third level. This takes more time and is more confusing to erect. However, if there is

a lot of scaffolding, this could be more economical when purchasing materials.

4.6.1 Securing Scaffolding to the Structure

Securing scaffold to the structure is the responsibility of the contractor and the scaf-folding engineer. When it is not specified, it is recommended to secure scaffoldingevery three levels. The attachment to the structure should be designed by a profes-sional, registered engineer. The attachment should consider the scaffolding movingtoward and away from the structure. Figure 4.10 shows what happens when scaffold-ing is not properly attached to the structure, considering wind and other forces besidesthe live load for which it was designed. Figure 4.11 indicates how a scaffold could beattached resisting both pushing toward the building and pulling from the building.

4.6.2 Hanging Scaffold

Hanging scaffolds are used more for cleaning and other exterior access to buildingswhen conventional lifts cannot be used due to reach or ground space. These types ofscaffolds are not typical to construction activities, but this is a great way to illustrateand calculate the different components necessary to design a similar system. Later,in Chapter 11, falsework removal with winches is discussed. This same concept canbe used when sizing the winches, wire rope, and counterweights on the winches.Figure 4.12 is a typical arrangement of how a scaffold may be suspended from abuilding or bridge. Two levels have been added to complicate the problem slightly.

FIGURE 4.10 Scaffold connection to building failure.


FIGURE 4.11 Scaffold connection to building example.

FIGURE 4.12 Suspended scaffold.

The hanging scaffolding in this example is two sets of supports on a bridge orbuilding. Each support consists of a counterweight (CW), a beam, a front support, anda wire rope cable. It is assumed that the wire rope is connected sufficiently to the beamand the scaffold. The object is to determine the weight of the whole scaffold. Applyhalf of that weight to a set of supports and design the counterweight and the beam.

Example 4.2 Design a hanging scaffold system similar to Figure 4.12 and with the

following characteristics:

A36 steel beams

Concrete counterweights (150 pcf)

IWRC 6 × 19 wire rope—improved plowed steel (IPS)

Scaffolding:

Light-duty Loading

32 ft long


5 ft wide

2 levels

Scaffold weighs 1200 lb

Factor of safety (scaffolding) 4.0 : 1.0

Factor of safety (wire rope) 6.0 : 1.0

Step 1: Determine the live load on the scaffold and add it to the 1200-lb scaffold

weight.

Live load, P = 2 levels (32 ft)(5 ft)(25 psf) = 8000 lb

Dead load, 1200 lb

Total scaffold weight, 9200 lb

Weight on one support system 9200 lb/2 sides = 4600 lb

Step 2: Determine the size of the IWRC 6 × 19 wire rope necessary to carry

half the scaffold. Table 4.2 has been provided to size the wire rope based on ulti-

mate breaking strength. Tables 12.5 and 12.6 will be different so that students of

temporary structures can experience different charts in the industry. The values in

Table 4.2 are for this book only. Wire rope manufacturers have specific charts for

their product.

TABLE 4.2 IWRC 6 × 19 Wire Rope Breaking Strength

Diameter (in) IPS (tons) XIP (tons) XXIP (tons)

1

42.94 3.4 —

3

86.56 7.55 —

1

211.5 13.3 8.30

9

1614.5 16.8 18.5

5

817.9 20.6 22.7

3

425.6 29.4 32.4

7

834.6 39.8 43.8

1 44.9 51.7 56.9

11

856.5 65 71.5

Minimum breaking force in tons, 4600 lb × 6.0 = 27,600 lb or 13.8 tons.

Table 4.2 has three types of wire rope. The quality of the wire rope increases from

left to right. It is recommended to use the lowest values in order to be safe. This is

also because the designer is not 100% sure the field engineer will order the specified

wire rope, if a higher quality is called out. According to Table 4.2, a minimum of9

16′′

wire rope needs to be used in order to withstand the 13.8-ton breaking strength load

with a 6 : 1 FOS.


FIGURE 4.13 FBD of single support beam.

Step 3:Determine the size of the counterweight. First draw an FBD of one support

system as shown in Figure 4.13.

P = 4.6 k

Sum moments about CW or R. Let’s sum moments about R.

(CW)(9′) = (4.6 k)(17′), CW = (4.6 × 17)∕9, CW = 8.7 k

Sum forces in the vertical direction to determine R.

+R − 8.7 k − 4.6 k = 0 R = 13.3 k

Remember, an FOS has not been applied to any of these forces.

Counterweight (CW) = 8.7 k, with an FOS of 4.0 : 1.0, multiply CW × 4.0, and

this is what the counterweight must weigh.

8.7 k × 4.0 = 34,800 lb

The unit weight of the concrete counterweight for this example is 150 pcf. In order

to determine the amount of concrete required for this weight, solve for the volume in

the following formula:

CW = 150 pcf × (bwh),where bwd is the volume in cubic feet

34,800 lb = 150 lb∕ft3 × volume, volume = 34,800 lb∕150 pcf = 232 ft3

Volume in cubic yards = 232∕27 ft3∕yd3 or 8.6 ft3 of concrete

The size of a deadman (counterweight), similar to Figure 4.14, if it were a perfect

cube, would be calculated by taking the cube root of 232 ft3.

3√232 ft3 = 6.14 ft (6′ − 2′′)

Step 4: Beam Design—Select the most economical beam while maintaining at

minimum of 1.25 : 1.0 beam stability ratio between the depth (d) and the flange width(bf). Draw an FBD, as shown in Figure 4.14, indicating the forces on a single beam.

Note: A beam stability ratio is a quick way to make sure the beam is not too unsta-

ble and risking flange buckling failure. However, this does not mean the beam will

not fail in lateral buckling. Later in the text, the student will learn how to check for

beam stability.


FIGURE 4.14 Counterweight.

FIGURE 4.15 Shear and moment diagram.

In step 3, the reactions were calculated. Using the same FBD, a shear and moment

diagram can be completed. The original forces (without an FOS) should be used for

this process and then an FOS will be added later.

The shear and moment diagram gives us a maximum moment of 78.25 ft-k.

Using A36 steel with a scaffold FOS of 4.0 : 1.0, the allowable bending stress on the

beam is 36,000 psi∕4.0 = 9000 psi. Using this steel, determine the most economical

(lightest) beam maintaining the 1.25 stability ratio between the depth (d) and the

flange width (bf ).Determine the minimum section modulus by using the bending stress formula and

solving for Sx.Sx = (78.25 ft-k × 12′′∕ft)∕9 ksi = 104.3 in3

List three wide flange beams with a minimum section modulus of 104.3 in3.

Table 4.3 shows three beams that meet the minimum section modulus.


TABLE 4.3 Three Beams That Meet Minimum Section Modulus

Beam Sx (in3) d (in) Bf (in) d∕bf (ratio) Check (OK or NG)

W10 × 100 112 11.1 10.34 1.07 OK

W12 × 79 107 12.38 12.08 1.02 OK

W14 × 74 112 14.17 10.07 1.40 NGa

aNG = Does not meet all the requirements.

The W12 × 79 is lighter than the W10 × 100 while still maintaining a 1.25 ratio

or lower. The W14 × 74 is the lightest and has the required section modulus, how-

ever, the beam is slightly too tall for its base and will probably have lateral stability

issues.

Select the W12 × 79.

CHAPTER 5

SOIL PROPERTIES AND SOILLOADING

Support of excavation in construction is used when excavations cannot legally or

physically support themselves and protect the workers inside. Trench or excavation

collapses are not the most frequent construction accident, but they occur in enough

cases to be one of the most deadly.

The loading on support of excavation systems has been a subject of study for

over a century, and many theories have been developed and accepted by engineers.

This chapter will review some soil mechanic basics and discuss and implement some

of the more common soil loading theories. A conservative approach is typically

taken in the following examples due to the sensitivity nature of excavation support

and safety.

5.1 SOIL PROPERTIES

Soils are classified into two main categories both for lab procedures and for tempo-

rary structure shoring design. The two classifications are coarse-grained soils (sands

and gravels) and fine-grained soils (silts and clays). This chapter will mention sev-

eral other characteristics that separate these two types of soils besides their particle

size. However, particle size is the most obvious difference. In soils classification,

one of the most elementary tests performed is the sieve analysis or gradation pro-

cess. In this test, samples are put through a series of different size screens (sieves)

to determine the amount of the sample that is retained or passes certain size screens.

The smaller grained the material, the more screens the particles pass; and the larger

grained the material, the more that is retained on the larger screens. The No. 200 sieve

has openings measuring 0.075mm or approximately 3/1000 of an inch. The amount

75

76 SOIL PROPERTIES AND SOIL LOADING

of material, by weight, that passes this sieve (or is retained by this sieve), determines

whether a material sample is a coarse-grained soil or a fine-grained soil.

Coarse-Grained Soils—Sands and gravels are defined as soils that have more than

50% of their particles retained on or above the No. 200 sieve (screen).

Fine-Grained Soils—Silts and clays are defined as soils that have more than 50%

of their particles pass through the No. 200 sieve (screen). The Unified Soils Classifi-

cation System is a commonly accepted testing system that includes information from

a particle size analysis for both fine-grained and coarse-grained soils and a plasticity

index test for fine-grained soils. A particle size analysis would place the samples in

their correct main category (coarse grained or fine grained), and the coarse material

would be separated between sands and gravels based on the amount of material pass-

ing or being retained by the No. 4 sieve. The No. 4 sieve measures 4.76mm (0.187 in,

which in fraction form is about3

16′′ ). The difference between sands and gravels is

at the 50% point of the No. 4 sieve. Less than 50% passing would be considered

gravels and more than 50% passing would be considered sand. Beyond the No. 4

sieve criteria, the amount of fines (silts and clays) are measured in the coarse-grained

materials to place them in an even more specific category.

Fine-grained soils are analyzed further for the plasticity index (PI). The PI is

derived from the Atterberg limit test, which measures the plastic limit and liquid

limit of a fine-grained soil in order to classify it as a clay or silt. The test sets limits

of how much water makes the soil solid, semisolid, plastic, or liquid. The percentage

difference in water between the plastic and liquid state is the PI. Table 5.1 shows how

soils are classified based on both the sieve analysis and the plasticity index.

TABLE 5.1 Soil Classification Example

Level 1

Classification

Level 2

Classification Fines

Soil

Symbol Soil Description

Coarse-grained

soils

Gravels Clean GW Well-graded gravel

GP Poorly graded gravel

With fines GM Silty gravel

GC Clayey gravel

Sands Clean SW Well-graded sand

SP Poorly graded sand

With fines SM Silty sand

SC Clayey sand

Fine-grained

soils

Silts and clays

LLa<50%Inorganic CL Lean clay

ML silt

Organic OL Organic clay/silt

Silts and clays

LLa 50% or

more

Inorganic CH Fat clay

MH Elastic silt

Organic OH Organic clay/silt

Highly organic Dark in color PT Peat

aLL = liquid limit.

ASTM Designation D-2487.

5.1 SOIL PROPERTIES 77

5.1.1 Standard Penetration Test and Log of Test Borings

Other tests will briefly be discussed later in this chapter, but the standard penetration

test (STP) and the log of test borings are always available to the engineer for design

and are typically included in the contract documents for construction and temporary

support design.

Every engineer/owner has a particular procedure he or she likes to follow for the

borings. State agencies tend to follow a fairly consistent format. The sample log in

Figure 5.1 is from a CalTrans project on the Yerba Buena Island used for the demo-

lition and construction of bridge structures after the new Bay Bridge was opened in

2013. Table 5.2 shows ranges for granular soil properties. State agency logs contain

a great deal of useful information for both the engineer and the contractor. Here is

some information that can be obtained by the log of test boring shown in Figure 5.1.

Bore hole number and date

Water table location/elevation and the date measured

Blow count: number of blows to produce 12 in of sampler penetration

Beginning elevation of bore (feet or meters above sea level)

Ending elevation of bore (feet or meters above sea level)

Size of sampler (diameter in inches or millimeters)

Description of material following the unified soil classification system

Dry unit weight of material (pcf)

Moisture content as a percentage

FIGURE 5.1 Log of test boring example.


TABLE 5.2 Properties of Granular Soils

Unit Weight (pcf)Apparent

Density

Relative

Density

(%)

SPT,

N60

(blows/ft)

Friction

Angle, 𝜙

(deg) Moist Submerged

Very loose 0–15 N60 < 5 <28 <100 <60

Loose 16–35 5 ≤ N60 < 10 28–30 95–125 55–65

Medium dense 36–65 10 ≤ N60 < 30 31–36 110–130 60–70

Dense 66–85 30 ≤ N60 < 50 37–41 110–140 65–85

Very dense 86–100 N60 ≥ 50 >41 >130 >75

Data provided by CalTrans.

5.1.2 Unit Weights above and below the Water Table

Soil can have different properties depending if they are in-place (in situ) properties or

below water table properties. This section discusses how the unit weights of different

materials are affected by their relationship to the water table. More emphasis will be

placed on unit weights below the water table since in-place (in situ) conditions are

analyzed with more standard methods and deal with moisture content fluctuations.

In this method, samples are dried in a kiln to remove the moisture before they are

analyzed further.

When Soil is below the water table, its unit weight is altered by how it is affected

when the water fills the voids. How the water affects the sample depends on whether

the sample is a sand, gravel, silt or clay.

Void Ratio and Determining Unit Weights below the Water Table The

void ratio is the ratio between the amounts of voids to the amount of solids in a

sample by volume after the water has been extracted. Void ratio is calculated with the

following formula:

e =VvVs

where e = void ratio

Vs = volume of the solids

Vv = volume of the voids

This ratio varies between fine-grained and coarse-grained materials. It should

also be understood how these two types of soils are affected by the intrusion of

groundwater. For instance, coarse-grained materials typically have more voids due

to their larger particle sizes. Therefore, when water is introduced into these voids,

the particles, predominantly sands and gravels, tend to float or become buoyant. This

water is known to be free draining. On the other hand, when water is introduced to

a fine-grained material, the water tends to get trapped within the particles and is not

free to travel through the material. This water is known to be non-free-draining.

An easy way to understand the difference between coarse-grained (CG) materials

and fine-grained (FG) materials is illustrated in the Table 5.3.

5.1 SOIL PROPERTIES 79

TABLE 5.3 Coarse Grained versus Fine-Grained Materialsa

Coarse Grained Fine Grained

Sand and gravels Silts and clays

Less than 50% passing No. 200 sieve More than 50% passing No. 200 sieve

Free draining below water table Non-free-draining below water table

Particle submerged below water table Particle saturated below water table

High 𝜙, low on zero C Low or zero 𝜙, high CStrength through mechanical bonding Strength through chemical bonding

a𝜙 is the Internal friction angle (degrees). C is cohesion (psf).

The shear strength of a particular material is dependent on how the particles within

the sample join together. In a coarse-grained (sand and gravel) material, the particles

lock together mechanically with their angular shapes. The angle of internal friction,

which is derived through lab testing, is typically between 20∘ and 45∘. Anythingbelow this range could be classified in the silts and clays category. Anything above

this range would tend to have other mechanisms helping the angle of internal friction

such as cementing capabilities. A fine-grained (silt and clay) soil relies on chemi-

cal bonding to join the particles together. Under a microscope, these particles would

tend to lay flat and parallel to one another, as opposed to sands and gravels, which

interlock in order to achieve shear strength. The silts and clays would not have a very

measurable angle of internal friction, but its cohesiveness would be appreciable.

Typically during soil analysis, a 1-ft3 sample is used in order to keep the ratios

in relative order in comparison to other samples. A volume of 1 ft3 one cubic foot

also keeps the math straightforward in comparison. In the following examples, unit

weights will be determined, given some initial soil information such as void ratio,

specific gravity, and the unit weight of freshwater, which will always be 62.4 pcf in

this text. (If salt water is being considered, unit weight is 64 pcf.)

Here are some common formulas that are used to determine unit weights based on

a soils ratio of solids to voids and its specific gravity. Some of these variables were

mentioned earlier.

e =VvVs

1 ft3 = Vs + Vv

Ws = Vs × SG × 𝛾water

Unit weight submerged (CG) = Ws − (𝛾water × Vs)

Unit weight saturated (FG) = Ws + ( 𝛾water × Vv)

whereWs is the weight of the solids or the weight of a dry sample assuming the air has

no weight; SG is the specific gravity of the material. Specific gravity is the relative

density of a solid material as it relates to water. Specific gravity of water is 1.0; and

𝛾water is the unit weight of water. Freshwater weighs 62.4 pcf and salt water weighs

64 pcf.


Example 5.1 Unit Weight of a Coarse-Grained MaterialCalculate the unit weight of a silty-sand below the water table that has the following

properties:

Void ratio of 0.55

Specific gravity = 2.70

Step 1: Determine the volume of the solids (Vs). Vs = 1∕1 + eVs = 1∕1.55 = 0.645 (64.5 % of this sample is solid material)

Therefore:

Vv = 1 − Vv, or Vv = 1 − 0.645 = 0.355 (this value is not needed to

complete solving this problem)

Step 2: Calculate the dry unit weight (Ws)

Ws = Water × SG × Vs, or Ws = 62.4 pcf × 2.70 × 0.645 = 108.7 pcf

Step 3: Calculate the submerged unit weight. Since this material is a coarse-

grained material (sand or gravel), the particles become submerged under

water and thus float in the water.

Step 4: Submerged unit weight = Ws − (water × Vs), orWs = 108.7 pcf

− (62.4 pcf × 0.645) = 68.45 pcf = 68.45 pcf

Notice how light the sample becomes when its particles are submerged in water.

For the same reason a rock is lighter when it is held under water, a submerged

sand/gravel becomes lighter.

Example 5.2 Unit Weight of a Fine-Grained MaterialCalculate the unit weight of a sandy-clay below the water table that has the following

properties:

Void ratio of 0.85

Specific gravity = 2.65

Step 1: Determine the volume of the solids (Vs). Vs = 1∕1 + e Vs = 1∕1.85 =0.541 (54.1% of this sample is solid material)

Therefore:

Vv = 1 − Vv, orVv = 1 − 0.541 = 0.459 (this value is needed in step 4)

Step 2: Calculate the dry unit weight (Ws)

Ws = Water × SG × Vs, or Ws = 62.4 pcf × 2.65 × 0.541 = 89.5 pcf

Step 3: Calculate the saturated unit weight. Since this material is a fine-grained

material (silts and clays), the sample becomes saturated under water, thus

making the soil heavier.

Step 4: Saturated unit weight = Ws + (water × Vv), orWs = 89.5 pcf +(62.4 pcf × 0.459) = 118.1 pcf

5.2 SOIL LOADING 81

Notice how heavy the sample gets because of the water being trapped between

the fine-grained particles. Anyone who has shoveled wet clay can attest to the weight

of the material as well as the stickiness. However, while the unit weight of clay is

heavier when saturated with water, the cohesiveness plays a role in allowing a clayey

material to stand on its own when a trench is being excavated. This is just the opposite

with a sand or gravel because sands and gravels tend to slope at the angle of repose.

5.1.3 Testing

There are many types of soil tests that can be performed for both permanent andtemporary designs. The Atterberg limits test was mentioned earlier and is just oneof many tests required for soil classification. Grain size tests were also mentioned asthe premier test for defining coarse and fine-grained soils. In addition, triaxial sheardevelops angle of internal friction values, and the pocket penetrometer test is used inthe field to determine in-place shear strength of soil so a competent person can classifythe trenched material and what type of sloping is required. This type of field testingis discussed at the end of this chapter. These and other tests are listed in Table 5.4.

5.2 SOIL LOADING

5.2.1 Soil Mechanics

Soil can be considered a liquid pressure, but measures are taken to consider the man-ner in which the particles interact. With a course-grained material, the particles worktogether mechanically; and the higher the angle of internal friction, the more shearstrength the soil will have. On the other hand, a fine-grained soil’s particles bondchemically, which creates shear strength through cohesion. In addition, the same twosoil classifications discussed above—fine grained and coarse grained—will be usedwhen determining the proper formulas.

Two conditions will be discussed when soil loading is considered: active and pas-sive pressure.

TABLE 5.4 In Situ Lab and Field Test Designations

Symbol Test

AL Atterberg limits

CA Corrosivity

CN Consolidation

MD Maximum density

DS Direct shear

R R value

PP Pocket penetrometer

SA Grain size

SE Sand equivalent

UC Unconfined compression

UU Unconsolidated, undrained triaxial

W Wash analysis


5.2.2 Active Soil Pressure and Coefficient

Active soil pressure is when the soil loads the structure in question. Most shoring

designs fall into this category because the purpose is to support an excavation or

trench with adequate support that resists collapse and minimizes deflections and

settlement.

The active soil coefficient is a multiplier used in the most common soil pressure

formulas to introduce the angle of internal friction into the estimated soil pressure.

The angle of internal friction is a very important element to soil pressure in Rankine

and other theories. The angle of internal friction produces reliable information that

is relevant to how soil loads a support system. This value, designated by the symbol

𝜙 can be obtained from soils reports and laboratory testing. For the active case, the

active soil coefficient can be determined as follows:

Ka =1 − sin 𝜑

1 + sin 𝜑

This value will always be less than 1.000. For more accurate calculations, the soil

coefficient values should be carried out to three decimal places (0.000). Anything less

could result in pressures off by up to 20 psf.

Table 5.5 shows estimated unit weights (𝛾) and applicable active soil coefficients

(Ka) for a variety of soil types based on their angle of internal frictions (𝜙). The anglesdecrease from gravels (highest) down to silts (lowest). These values can be found in

soils reports and field testing reports, and some of these soil types and associated

values are found in Table 5.5.

TABLE 5.5 Simplified Typical Soil Values

Soil

Classification

𝜙 Friction

Angle of

the Soil

Density

or

Consistency

𝛾 Soil

Unit Weight

(pcf)

Ka Coefficient

of Active Earth

Pressure

Gravel,

gravel–sand

mixture,

coarse sand

41

34

29

Dense

Medium dense

Loose

130

120

90

0.21

0.28

0.35

Medium sand 36

31

27

Dense

Medium dense

Loose

117

110

90

0.26

0.32

0.38

Fine sand 31

27

25

Dense

Medium dense

Loose

117

100

85

0.32

0.38

0.41

Fine silty sand,

sandy silt

29

27

25

Dense

Medium dense

Loose

117

100

85

0.35

0.38

0.41

Silt 27

25

23

Dense

Medium dense

Loose

120

110

85

0.38

0.41

0.44

5.2 SOIL LOADING 83

Passive Soil Pressure and Coefficient Passive soil pressure is when astructure loads the soil. For example, this could occur when a retaining wall beingloaded from behind by soil subsequently pushes on the soil in front of the wall.Another case would be when an anchor plate or block is supporting a tieback systemand the force from the tieback is pushing the anchor/plate against the soil (structureloading soil).

As with the active case, the passive has a soil coefficient that is derived from theangle of internal friction. The following value is the reciprocal of the active soil coef-ficient. Because the passive soil coefficient is the reciprocal, the value will always begreater than 1.000 and sometimes as high as 5.000.

Kp =1 + sin 𝜑

1 − sin 𝜑

5.2.3 Soil Pressure Theories

Several theories have been developed over the years for soil pressure analysis. Rank-ine, Boussinesq, Coulomb, Terzagie, and others have derived several theories thatsupport how soil loads its support system. In the effort to simplify some of the tem-porary structure concepts in this text, the author has condensed the most commonpractices in order to give the student a solid, but simplified, understanding of soilloading. This is not to say that very detailed analyses should not be performed by aregistered engineer, but the majority of our project cases can be solved using acceptedtheories. Test results on actual shoring systems have produced evidence that the mostcommon soil pressure theories coincide with practical results within 20%, regard-less of the method used. For this reason, the student should rely on future employersto dictate the methods used in determining soil pressures on structures. In addition,standard factors of safety should be used on a case-by-case basis.

Rankine Theory Rankine theory is a great method for a student of engineeringbecause it covers all the soil properties, and one can make sense of how these prop-erties apply to the amount of soil pressure generated by the conditions. This textwill start with the Rankine theory because it employs all the soil property variablesmentioned earlier: angle of internal friction, unit weight, and cohesion. Later in thischapter, other theories will be discussed that categorize cases into soil types and sup-port conditions. Also, asmentioned before, Rankine theory can be used for both activeand passive soil pressure cases.

Rankine Theory for Active Soil Pressure When soil is loading a structure inquestion, the pressure it develops based on Rankine at any depth of h is as follows:

Pa = γhKa − 2c√Ka

where Pa = active pressure at any given depth of h (psf)𝛾 = unit weight of the soil (pcf)

h = depth at which the pressure is being measured (ft)

Ka = Active soil coefficient; Ka = 1 − sin 𝜑∕1 + sin 𝜑

c = cohesion (psf)


The first part of the formula (𝛾hKa) takes into account the weight of the soils

and how deep the excavation or trench is. The soil coefficient is a multiplier in this

case and greatly affects the outcome of the pressure value. As mentioned before,

the active soil coefficient is always less than 1.000. Therefore, anything less than

1.000 is going to lower the value after taking into consideration the unit weight

and the depth. The higher the degree of internal friction, the lower the value of Ka,thus the lower pressure would result. In contrast the lower the friction degree, the

higher the coefficient value is closer to 1.000, thus the higher the pressure will be

from the soil. This occurs because the smaller the friction value the less shear strength

the soil has.

The second part of the formula (−2c√Ka) takes into account the amount of cohe-

sion (clay) the soil contains. Notice the negative sign. This value is subtracted from

the first part of the formula. Therefore, if there is cohesion in the soil, the soil pressure

will be lower. The clay causing the cohesion helps strengthen the soil, thus reducing

the pressure. If there is no cohesion to the soil, then this portion of the formula goes

to zero, and the pressure only comes from the unit weight, the coefficient of friction,

and the depth.

One thing to notice about this method is if the depth is 0 ft (ground level) and there

is cohesion in the soil, the outcome pressure can be a negative pressure. This means

that close to the top of the excavation, the soil is supporting itself. This should be

made clearer in the examples to follow.

For an active soil case, when the soil unit weight is unknown, theCalTrans Trench-ing and Shoring Manual require the engineer to use 115 pcf.

The slip plane of the active case is the measured angle from the vertical of the

structure to the plane of failure from the bottom of the excavation to the surface. This

angle is calculated as follows:

𝛼 = Slip plane angle = 45∘ − 𝜑

2

where 𝛼 = angle in degrees

𝜑 = angle of internal friction

Rankine Theory for Passive Soil Pressure In the case where there is passive

pressure and the structure is loading the soil, the formula is similar in its variables

but very different in its outcome:

Pp = γhKp + 2c√Kp

where Pp = passive pressure at any given depth of h (psf)Kp = passive soil coefficient, Kp = 1 + sin φ∕1 − sin φ

The first and most obvious difference is that cohesion is added to the pressure due

to depth, unit weight, and friction. When there is no cohesive value, this does not add

any strength to the soil. However, a cohesive soil that does have cohesion will add

a great deal of soil resistance. After all, passive pressure is basically soil resistance

against the structure that is pushing against it. Passive soil pressure will always be a

5.2 SOIL LOADING 85

positive pressure because the clay strength is always added and not subtracted likeactive pressure.

The slip plane of the passive case is the measured angle from the vertical of thestructure to the plane of failure from the bottom of the structure to the surface. Thisangle is calculated as follows:

𝛽 = Slip plane angle = 45∘ + 𝜑

2

where 𝛽 = angle in degrees

𝜑 = angle of internal friction

𝛼 = angle in degrees

5.2.4 Soil Pressure Examples Using Rankine Theory

Rankine Theory for Active Soil Pressure (Pa ) As already mentioned, theRankine theory is a great basis for soil pressure education because it considers theunit weight of the soil, the soil coefficient, the depth of excavation, and how cohesivethe soil might be. The following formula is based on the Rankine theory for activesoil pressure (Pa).

Pa = 𝛾hKa − 2c√Ka

Figure 5.2 shows diagrams of active soil pressure for granular and cohesive soils.

Rankine Theory for Passive Soil Pressure (Pp ) The following formula isused when calculating passive soil pressure using the Rankine theory:

Pp = 𝛾hKp + 2c√Kp

where 𝛾in situ = unit weight of soil (pcf)

h = depth of load from the surface

Kp = passive soil coefficient

FIGURE 5.2 Active soil pressure diagrams for granular and cohesive soils.


FIGURE 5.3 Passive soil pressure diagram.

Figure 5.3 provides the diagram of passive soil pressure.

Kp =1 + sin 𝜑

1 − sin 𝜑

where Φ = angle of internal friction:

Slip plane = 𝛽 = 45∘ + 𝜑

2

Example 5.3 Passive Soil Pressure Supporting a Steel PlateWe are supporting an SOE (support of excavation) with a tieback to a deadman, steel

plate (see Figures 5.4 and 5.5). The following describes the soil properties and the

FIGURE 5.4 Passive pressure diagram against plate.

5.2 SOIL LOADING 87

FIGURE 5.5 Trapezoidal pressure diagram on plate.

loading conditions. Determine the Pmax at the top and bottom of the plate and the

total capacity of the plate:

γin situ = 110 pcf

c = 250 psf

Φ = 18∘

The 8-ft-wide × 6-ft-high steel plate is buried 10 ft deep to the top of the plate.

Step 1: Calculate Kp = 1 + sin 18∕1 − sin 18

Kp = 1.894

Step 2: Calculate the pressure at the top of the plate.

Top Pp = 110 pcf × 10′ × 1.894 + 2 × 250 psf ×√1.894 = 2772 psf

Step 3: Calculate the pressure at the bottom of the plate.

Bottom Pp = 110 pcf × 16′ × 1.894 + 2 × 250 psf ×√1.894 =

4022 psf

Plate capacity =[(Ptop + Pbtm

2

)× plate w × plate h

]/FOS

1.5

(2771 + 4021)∕2 × 8′ × 6′= 108, 672 lb (54.3 tons)

Example 5.4 Intersecting Slip PlanesIn order to determine a safe distance behind an excavation for equipment so that the

equipment surcharge load does not influence the shoring load, calculate the horizontal

distances created by both passive and active slip planes. The goal would be tomaintain

a zero or positive distance between X1 and X2 (see Figure 5.6).Given information should be the depth of excavation, the two angles 𝛽 and 𝛼, and

the depth to the bottom of the anchor plate. From this point, the calculation is strictly

geometry and trigonometry:

D = 30 ft to bottom of excavation

d = 20 ft to bottom of plate


FIGURE 5.6 Failure planes for active and passive.

𝛽 = 54∘

𝛼 = 45 − 18

2= 36∘

X1 = tan 36 × 20 ft = 14.53′

X2 = tan 54 × 30 ft = 41.29′

X1+X2 = 14.53 + 41.29 = 55.82 (safe distance for equipment behind shoring wall)

Active Slip Plane with Soil Anchors When soil anchors are used to tie back a

support of excavation, the length of the anchor is critical to its capacity. The wedge

of earth between the failure plane and the vertical shoring face cannot be relied upon

for support because this soil is already unstable. Therefore, once the soil anchor length

is determined, the wedge width should be added to this minimum length in order to

achieve the full drilled depth of the anchor (see Figure 5.7).

FIGURE 5.7 Active slip plane.

5.2 SOIL LOADING 89

Example 5.5 Active Soil Pressure on Sheet Pile Wall with a Granular SoilWe are supporting an SOE for a 30-ft-deep excavation with a sheet pile wall. The

following describes the soil properties and loading conditions. Determine the Pmax at

the bottom of the excavation. Refer to Figure 5.8 for the pressure diagram:

𝛾in situ = 120 pcf

c = 0 psf

Φ = 35∘

D = 30-ft-deep excavation

Step 1: Calculate Ka = 1 − sin 35∕1 + sin 35.

Ka = 0.271

Step 2: Calculate the pressure at the bottom of the excavation.

Bottom Pa = 120 pcf × 30′ × 0.271 + 2 × 0 psf ×√0.271 =

975.6 psf

𝛼 = 45 − 35

2= 27.5∘

FIGURE 5.8 Active soil pressure diagram.

Example 5.6 Active Soil Pressure on Sheet Pile Wall of Sandy ClayWe are supporting an SOE for a 30-ft-deep excavation with a sheet pile wall. The

following describes the soil properties and loading conditions. Determine the Pmax at

the bottom of the excavation. Refer to Figure 5.9 for pressure diagram.

𝛾in situ = 130 pcf

c = 500 psf

Φ = 10∘

D = 30-ft-deep excavation


FIGURE 5.9 Active soil pressure diagram.

Step 1: Calculate Ka = 1– sin 10∕1 + sin 10.

Ka = 0.704

Step 2: Since there is cohesion, calculate the pressure at the top of the excavation.

Bottom Pa = 130 pcf × 0′ × 0.704 + 2 × 500 psf ×√0.704 =

839 psf

Step 3: Now calculate the pressure at the bottom of the excavation.

Bottom Pa = 130 pcf × 30′ × 0.704 + 2 × 500 psf ×√0.704 =

1907 psf

Step 4: Calculate the distance from the surface to where the pressure crosses 0

psf (the vertical line).

Step 5: Draw two similar triangles with the upper and lower portion of the pres-

sure diagram (see Figure 5.10). Calculate the unknown vertical distance y.

y

839= 30′

2746y = 9.2 ft

From the top of the excavation to 9.2 ft below the surface, theoretically

the soil is supporting itself.

𝛼 = 45 − 10

2= 40∘

FIGURE 5.10 Similar triangles for y dimension calculations.

5.2 SOIL LOADING 91

5.2.5 Soil Pressures Using State and Federal Department Standards

Curved Failure Planes We have discussed Rankine’s theory in depth thus far.Before Rankine (1857), Coulomb had developed his own theories that are still rec-ognized today. However, it was not until 1943 that Terzaghi brought a better under-standing to how the original earth pressure theories work. Coulomb’s and Rankine’s

theories assumed a wedge-shaped failure plane, whereas Terzaghi introduced a morelogarithmic failure plane shape, shown in Figure 5.11. This is the same failure planediscussed under Rankine’s theory described as a slip plane.

Field experiments have shown that Rankine and Coulomb theories are very accu-rate when they are applied to active soil pressure cases but only accurate in passivecases with clean dry sand.

As mentioned earlier, soil pressures can also be developed using other theories.The following theories have been developed by engineers and adopted by state and

federal agencies. Their diagrams appear to have been simplified; however, field test-ing validates their accuracy. These theories are classified by soil type and supportcondition. The soil types are granular and fine grained. The support conditions areunrestrained (cantilevered) and restrained (single support and multiple supports). Theunrestrained systems rely of passive soil resistance below the bottom of the excava-tion as they push back against the structure. In addition, the bending capacity of thestructure itself is also a large component of the strength of the system. These systemsare usually not effective with depths greater than 15–18 ft. The restrained systems

typically use the same vertical shoring members, but instead of relying on passivesoil support, they rely on walers (horizontal beams) and struts (compression mem-bers) or tiebacks (tension members) to complete the system. With restrained systems,contractors can support excavations over 60 ft in depth safely and economically.

Both types of support systems count on the vertical members toeing into the sub-grade for some distance. However, this is definitely more necessary with the unre-strained system since this is where it gets its passive soil resistance.

Earlier in this chapter, the defining line between granular soil and fine-grained soils

was discussed. In addition, the difference between sands and gravels was defined.

FIGURE 5.11 Curved failure plane.


FIGURE 5.12 Triangular pressure diagram.

As far as these theories go, sands and gravels are combined into the same theory. Lat-

eral pressures for both active and passive cases are determined by the soil properties

and the type of shoring system.

The triangular pressure diagram (Figure 5.12) that varies with depth has a resultant

one third of the depth from the bottom of excavation. The maximum pressure in this

case is at the bottom of excavation. Horizontal stress is calculated by

Horizontal stress (𝜎h) = 𝛾hK

Resultant = 𝜎hh∕2

Soil typically is classified as a liquid even though hydrostatic groundwater pressure

is added to the lateral soil pressure. When soil pressure diagrams are added together,

the normal triangular shape becomes a series of triangular and rectangular shapes

as shown in Figure 5.13, which shows soil and water pressure, both separately and

combined.

Standard Pressure Formulas Simplified This text has categorized just a few

shoring conditions that cover the majority of common situations while avoiding very

complex situations. Shoring conditions that are more extreme than the examples used

in this text should be analyzed and designed by a registered professional engineer.

The shoring embedment below the bottom of excavation has been simplified to a

FIGURE 5.13 Combined soil and water pressure.

5.2 SOIL LOADING 93

TABLE 5.6 Most Common Soil Pressure Formulas

Category No. Description Formula

1 Cantilevered and single brace, granular soil Pa = Ka𝛾H2 Cantilevered and single brace, cohesive Pa = Ka𝛾H − 2c3 Multiple brace, granular or cohesive soil Pa = 0.65Ka𝛾H

where Pa = maximum active soil pressure

0.65 = reducing multiplier

Ka = active soil coefficient

𝛾 = soil unit weight

H = depth to bottom of excavation

C = shear strength of cohesive soil (psf)

ratio to avoid complicating the loading diagram. The categories selected are listed in

Table 5.6.Some assumptions are made when using these theories. The main assumptions are

as follows:

The wall and the soil adhere to each other through cohesion and friction.

The ground elevation is level to the top of the shoring (𝛽 = 0∘).The failure wedge is a function of the soil’s internal friction angle.

The lateral pressure acts perpendicular to the shoring.

The soil types that follow either fall into the granular category or the cohesive cat-egory. These differences, combined with the support type, make up the differences

between the theories. It should be noted that the granular-type soils are represented

well between the Rankine and the Coulomb theories. However, when cohesive soilis present, a modified Rankine theory is better suited. The credit for this modifica-

tion, which was shown previously in this chapter, goes to Bell in 1952. The original

Rankine formula, also shown above, uses this modification in the second part of theformula (±2cKa or Kp). It should be repeated that this is how the top of the excavation

in an active soil pressure case can be a negative number. This formula modification

also does not take into account the effects of groundwater and hydrostatic pressure.Lateral soil pressure acting on a wall with a height of h should never be less than

25% of the vertical effective stress, which is 𝜎 = 𝛾h. Vertical stress is equal to unit

weight of soil (𝛾) times the depth (h). The only way an engineer can use a lesser

value would be if extensive laboratory testing can verify that cohesion is higher thanspecified.

Wall friction is the presence of bonding between the shoring and the soil.

Coulomb’s theories diminish in credibility because the theory does not considerthe effects of wall friction. In fact, the theories in this text ignore wall friction

calculations and use additional factors of safety to compensate for the fact that wall

friction varies with soil type.Figure 5.14 shows the combination between active and passive soil pressure when

a curved failure plane is considered.


FIGURE 5.14 Passive and active pressures with curved failure.

FIGURE 5.15 Cantilevered shoring system.

Cantilevered Shoring Systems Shoring systems using either sheet piling or

soldier beams can typically cantilever (without support) 15–18 ft in depth. Beyond

these depths, the engineer should consider tiebacks or a strutting system as mentioned

in the next section. Cantilevered systems (see Figure 5.15) rely on the strength of the

soil below the subgrade as well as the strength of the sheet piling resisting bending.

The toe of the shoring system is the depth the vertical member is embedded below the

bottom of excavation. In cantilevered systems, this distance can sometimes be 1.5–2.0

times the length of vertical shore supporting the excavation depth. In other words,

if an excavation is 16 ft deep, the vertical shoring component (sheet pile of soldier

beam) could be 40 ft [(16 × 1.5) + 16] to 48 ft [(16 × 2.0) + 16] long, depending on

the required embedment toe.

Embedment calculations can be complicated; therefore, many professional engi-

neers use computer software to help calculate these values. This text will attempt to

simplify this process. As mentioned previously, one way to simplify this distance is

to use a multiplier from 1.5 to 2.0. This range can determine this dimension safely

in most cases. This is especially common when an engineer is doing a projected cost

estimate and is only trying to get close to the quantity of shoring required. The con-

struction manager, in many cases, is trying to accomplish just that in order to get a

5.2 SOIL LOADING 95

rough, but conservative, estimate of materials so that an educated cost estimate canbe achieved.

The example in Figure 5.15 requires a distance ofD in order to calculate the activeand passive pressures at the different depths. The pressure most important for calcu-lating shoring component sizes is the maximum active soil pressure at the bottomof the excavation Pa = Ka𝛾H. This value can be used for pressure calculations ingranular soils

1. Cantilevered and Single Brace System in Granular Soil Shoring in granularsoil (both dense and loose) should be designed for a rectangular load where Pa =Ka𝛾H. Figure 5.16 illustrates this soil pressure graphically, whereP is lateral pressure,Ka is active soil coefficient, 𝛾 is the soil unit weight, and H is the depth of excavation.

2. Cantilevered and Single Brace System in Cohesive Soil When a cantileveredsystem is supported in a cohesive material such as clay, the soil load can be repre-sented by Pa = Ka𝛾H − 2c. Figure 5.17 shows this pressure diagram, and the maxi-mum soil pressure is shown by the largest pressure area in the middle of the diagram.The upper portion of the pressure can be negative due to subtracting the cohesion(−2c). This is usually just assumed to be zero pressure because there can’t actuallybe zero pressure.

Supported Shoring Systems Once it is determined that a system is too deepto support itself without walers, tiebacks, and/or struts, the soil loading conditions

FIGURE 5.16 Cantilevered system in granular soil.

FIGURE 5.17 Cantilevered and single brace system in cohesive soil.


FIGURE 5.18 Supported shoring system components.

change and so does the support system. When more than one brace is required,

another formula can assist the engineer in determining an accurate lateral soil

pressure. Bracing consists of walers, struts, and/or tiebacks. These components will

be briefly introduced here, then actually designed in Chapters 6 and 7. Figure 5.18

shows how these components correspond to a supported shoring system.

Walers: Horizontal support member that supports the vertical shoring (sheet piling

or soldier beams) or transfers the shoring load to tiebacks or struts. Figure 5.19 is a

photo of a double steel channel waler using tiebacks. The double waler is being sup-

ported by “lookout” brackets. These brackets support the waler until the soil pressure

on the system supports them.

Tiebacks: Tension rods or strands that support walers or soldier beams and trans-

fers load into the supporting soil. A tieback must be long enough to extend through

the failure wedge of soil and adequately into the competent earth. Tieback design will

be mentioned later.

FIGURE 5.19 Double steel channel waler with tiebacks.

5.2 SOIL LOADING 97

Struts: Compression pipes, beams or timbers that support walers and transfer loads

to the other side of the excavation. Since struts act like columns, a high slenderness

ratio will limit the struts buckling capacity. In this case, tiebacks should be considered

or the struts may have to be supported vertically as well.

3. Multiple Brace, Granular or Cohesive Soil The third system described in this

section is the system that is deep enough to requiremore than one level of support. The

maximum pressure for this situation is reduced by 65% because of the number of sup-

ports (>1). This assumes water is not present. If water is present, Pw = 62.4 pcf × H,should be added. Also, the soil unit weight should be the saturated or the submerged

unit weight. The pressure for a multiple supported system with either granular or

cohesive soil is calculated as Pa = 0.65Ka𝛾H. Figure 5.20 illustrates this pressure

diagram.

Figure 5.21 is a photo of a multiple supported excavation in San Jose, California.

The system had three levels of supports using walers and struts. Each level from top

to bottom was installed as the excavation proceeds downward.

FIGURE 5.20 Rectanglar pressure diagram.

FIGURE 5.21 Multilevel supported excavation.


Effects of the Water Table on Loading Water table changes the unit weight

of the soil, which in some cases can lower the unit weight and in some cases increase

the unit weight. Water can also change the shoring system selection by forcing the

designer to use a water-tight system where, in some cases, another, less expensive,

system could have supported the load. Typically, an excavation has to be dry in order

to perform the work inside such as concrete placement. Pure hydrostatic water pres-

sure must be added to the soil loads in a free draining case where granular soil is

dominant. In a cohesive soil case, the soil unit weight represents a saturated unit

weight, which is going to be much heavier than its dry weight, therefore, the 𝛾 value

takes this into account as far as pressure is concerned.

Surcharge Loads Surcharge loads are typically from existing buildings/struc-

tures, construction equipment, traffic, or materials. This text will focus on construc-

tion equipment surcharge loads. Standard practice uses either a theoretical value or

an actual weight when a specific piece of equipment is specified.

Theoretical surcharge loads are uniform loads predetermined by an engineering

counsel and covers standard equipment up to a certain weight. Equipment that is

heavier than the standard would have to fall into the actual weight category described

below. The minimum lateral construction surcharge load is 72 psf at 10 ft in depth

below the shoring system regardless if the system is subject to any surcharge loading

during construction. This would be the equivalent of soil that weights 109 pcf and

an active soil coefficient of 0.333 for a depth of 2 ft. For surcharge load to affect the

lateral loading the surcharge load should be within the active soil wedge. Figure 5.22

illustrates a surcharge load (Q) and how it relates graphically to a lateral load in the

Hs zone:

P horizontal = KaQ

where Q = equipment load per square foot within the limits of the active failure

wedge

Ka = active soil coefficient

Phorizontal = lateral pressure on the shoring due to the surcharge load

When equipment weight is being estimated, one can apply a series of uni-

form loads that change as the depths increase. This typical surcharge loading

FIGURE 5.22 Surcharge loading.

5.2 SOIL LOADING 99

FIGURE 5.23 Standard surcharge loads by depth.

FIGURE 5.24 Equipment surcharge load on an excavation.

is recommended when standard, small equipment is planned for construction.

Figure 5.23 shows the surcharge values decreasing as the depth of excavation

increases. The values are higher closer to the surface and decrease as the excavation

gets deeper. Figure 5.24 shows a photo of a long-reach excavator loading trucks

next to a sheet pile support of excavation. This excavation is over 30 ft deep, so the

excavator must position right up against the sheet piles.

Actual Equipment Weight Method There are many cases where a contractor

designs a shoring system to support the lateral soil loads and an additional surcharge

load for a particular crane or excavator, which is going to be the largest and heaviest


used during the life of the shoring system in place. As an example, if a contractor

plans to use a 150,000-lb crane with a 25′ × 18′ base dimension, this would equate to

qs = 333 psf, where qs is the actual surcharge load in psf. In practice, the surcharge

causes a lateral pressure of P = Ka × qs = 0.33 × 333 (33% of the load). Therefore,

the designer would add a rectangular lateral load of 111 psf for the depth of the exca-

vation. If this force is 72 psf or less, a minimum force of 72 psf should be used. In

no case should a surcharge load be less than 72 psf. This will be explained further in

the next section.

Equipment weight, 150,000 lb

Base dimensions, 25′ × 18′

qs = 333 psf

Surcharge load, 333 × 0.33 = 111 psf

The deeper the excavation is, the less effect the equipment weight has to the relative

depth. There is theoretically a depth where the surcharge load has no effect. However,

as discussed in the previous section, a minimum of 72 psf is still used.

Boussinesq Loads Boussinesq developed three type of surcharge loads; the strip

load, the line load, and the point load (Figure 5.25). These three types of surcharge

loads were arranged for three particular cases. These cases are as follows:

Strip load—Highways, roads, or railroads parallel to the shoring

Line load—continuous wall or footing parallel to the shoring

Point load—outriggers from crane or pump

Table 5.7 shows three different surcharge loads and their totals when combined;

72 psf is the minimum surcharge load that can be used if equipment will be working

within the limits of an excavation support.

OSHA Trench Classifications by Soil Type This text focuses on designed shoring

systems using sheet piling or soldier beams. In the case of trenching for utilities and

other horizontal underground applications, state and federal OSHA regulations have

FIGURE 5.25 Surcharge loads using Boussinesq.

5.2 SOIL LOADING 101

TABLE 5.7 Surcharge Lateral Pressure from Equipment

Depth (ft) Q = 100 (psf) Q = 200 (psf) Q = 300 (psf) Sum (psf)

0.1 1.9 0.3 1.7 72min

1 17.9 3.0 17.1 72min

2 30.2 5.8 33.8 72min

4 35.7 10.1 63.7 109.5

6 29.5 12.3 87.1 128.9

8 21.9 12.7 103.3 137.9

10 15.9 11.9 112.6 140.4

12 11.5 10.5 116.4 138.4

14 8.5 9.0 116.1 133.6

16 6.3 7.6 112.9 126.8

Source: Data from Cal Trans Trenching and Shoring Manual (2011).

TABLE 5.8 Soil Properties by Type

Soil Type Unconfined Compressive Strength (TSF)a Minimum Slope of Trench Wall

A 1.5 3/4:1

B 0.5 1:1

C N/A 1 1/2:1 or greater

aTons per square foot.

created guidelines. These guidelines classify soil types by type A, B, and C. Most

jurisdictions require contractors to employe a “competent” person in soil identifica-

tion. This person can then determine either sloping criteria or shoring types. These

shoring types would be in the category of shielding, speed shoring, or a combina-

tion of both. Table 5.8 briefly describes the characteristics required for a soil to be

a type A, B, or C.

The competent person on the project must go through a one-day (minimum) train-

ing course in order to be able to identify soil into one of the three categories listed

in the table. This person is responsible for deciding on the type of shoring or sloping

systems to be used and if engineering is required. The competent person’s title should

not be confused with a licensed, registered engineer. If engineering is required and the

competent person does not meet the requirements of a licensed, registered engineer,

that person should employ the proper person.

Other Project Examples The Figures 5.26–5.29 show some examples of work

that involved support of excavation and soil type in addition to what was already

discussed in this chapter. Figures 5.26 and 5.27 are examples of cemented soils in a

cofferdam where sands and gravels were expected. Not only did the shoring installa-

tion suffer due to the unexpected soils conditions, Figure 5.29 shows how the exca-

vation methods employed were very slow. A small track excavator was used with a

1000-lb breaker and the material was loaded into a skip box for removal from the

cofferdam.


FIGURE 5.26 Soil conditions in a sheet pile cofferdam.

FIGURE 5.27 Methods of excavation for cemented soils.

Figure 5.28 shows a sheet pile cofferdam in a river with external walers. This

system will be looked into further in Chapter 7. Dewatering pumps are shown in the

background. It took two 4-in submersibles to keep up with the dewatering efforts due

to split sheets and sheets that were dislodged from their interlocks.

5.2 SOIL LOADING 103

FIGURE 5.28 Sheet pile cofferdam with external walers.

FIGURE 5.29 Slide rail system.

Figure 5.29 shows a slide rail system used in San Francisco to shore bridge pier

footings. This type of system is preengineered and only works with its own compo-

nents. Rental companies have systems like this available to rent and usually provide

any engineering information required to submit to the owner/engineer for approval.

CHAPTER 6

SOLDIER BEAM, LAGGING,AND TIEBACKS

Now that soil properties and loading conditions have been covered, they will be incor-porated with the study of two common support of excavation systems: soldier beam,lagging, and tiebacks in this chapter and sheet pile, walers (sometimes referred to aswales), and struts in the next chapter.

As with the soil loading, state and federal department trenching and shoringmanuals are used in this text for shoring design. There are some cases where railroadand U.S. Navy standards may be more applicable. However, when not specificallystated, state and federal departments will be the standard. The examples in this chapterand the next have been fairly simplified so that the main concepts are not lost in thefiner details.

6.1 SYSTEM DESCRIPTION AND UNITS OF MEASURE

6.1.1 Beams/Piles

Beams and piles are the foundation of the soldier beam and lagging system. Theyare the components that embed from original ground to below the subgrade of theexcavation, providing anchor support for the system. Beams and piles are typicallyeither drilled then set and grouted or driven. When there are sensitive environmen-tal concerns, drilling seems to be the method of choice. However, when produc-tion is the only concern, driving seems to be most economical. Beams and pilesare placed by the individual beam as a unit of each when estimating and trackingcosts.

The beams and piles support the lagging, and their flangesmake direct contact withthe soil. However, for design purposes, the flanges of the beams support the laggingmaterial, which is what supports the soil with which it makes direct contact.

104

6.2 MATERIALS 105

6.1.2 Lagging

Lagging is the material in this system that makes direct contact with the soil. Laggingis typically wood planking or steel plates.Wood can be expensive to purchase initiallybut, if reusable, can be economical over several uses. Steel plates, per unit, are evenmore expensive thanwood lagging but have amuch higher probability of being reusedmultiple times. When estimating or tracking cost of this work, it is most common toquantify by the square foot (SF or ft2). Wood prices can range from $0.60 to $1.20 perboard foot, and steel plate prices can range from $0.45 to $0.75 per pound. Owningeither of these materials, especially steel, is paramount to a company’s success if thistype of work is self-performed frequently.

Steel plates can be rented from shoring and trench plate rental companies but canbe very expensive over the span of the rental period.

6.1.3 Tiebacks

Tiebacks, soil nails, and soil anchors are three different types of anchors usedto support shoring systems. They range in size from small anchors (soil nails)to medium-size anchors (tiebacks) to larger ties (soil anchors). The larger, highercapacity the anchor, the more expensive the anchor and larger the shoring systemcomponents are. Alternatively, the smaller the anchor, the lighter the accompanyingsystem is. The smaller the size of the system typically translates to a lower cost.However, if the system requires a large quantity of expensive ties, the cost increases.Therefore, ties are quantified by size for estimating or cost tracking purposes.

6.2 MATERIALS

6.2.1 Steel AISC

For temporary steel construction, theAmerican Institute of Steel Construction (AISC)is the governing standards, except when some jurisdictions may require a higher fac-tor of safety. The steel most commonly used for temporary structures is ASTM A36steel, which has an ultimate strength of 58,000–66,000 psi and yield strength (Fy)of 36,000 psi. For rolled steel sections, A36 grade steel has the following allowablestress values:

Fv = 0.4Fy = (0.4)36 ksi = 14.4 ksi

E = 29,000,000–30,000,000 psi (almost always 29,000,000 psi)

Fb (noncompact sections) = 0.6Fy, (0.6)36 ksi = 21.6 ksi

Fb (partially compact) between 0.6Fy and 0.66,Fy = (21.6 − 23.76 ksi)

Fb (compact sections) 0.66Fy = (0.66)36 ksi = 23.76 ksi

No matter which method for bending is used, the lowest value is 21.6 ksi. There-fore, when in doubt, using 21.6 ksi allowable bending stress is recommended. Differ-entFb values can be requiredwhen the compression flange of the beam is unsupportedor not fully supported. Flange buckling stress will be covered later in this text.

106 SOLDIER BEAM, LAGGING, AND TIEBACKS

6.2.2 Wood Species—National Design Specifications (NDS)

for Wood Construction

Wood lagging is used quite often in soldier beam and lagging systems. In most cases,

standard Douglas fir–Larch lumber is used. However, pressure-treated wood can be

used when the system is installed for longer periods of time or even for permanent

construction. When a species of lumber is not specified, the following values are the

maximum allowable stress values this book will use in shoring applications:

Fcll = 480,000 (L∕D)2 psi; not to exceed 1600 psi

Fb = 1800 psi > 8′′ nominal depth

Fb = 1500 psi ≤ 8′′ nominal depth

Ft (direct tension) = 1200 psi

Fc,perp = 450 psi

Vhor shear = 140 psi

E = 1,600,000 psi

Contractors commonly know what material will be used on a system before con-

struction. Therefore, they should use values for that specific material as listed in

the NDS.

Some agencies, such as railroads, may use slightly different values than those used

for highway transportation. In these cases, the values are sometimes lower than the

standard for the NDS. For example, railroad only allows 1710 psi for bending instead

of 1800 psi. Deflection requirements are not as strict with shoring as in, say, con-

crete formwork, which usually leaves a permanent product (concrete wall) exposed

to view, thus requiring tighter tolerances. The soldier beam and lagging shoring wall

can deflect without being out of specification.

Short-Term Loading Soldier beam and lagging systems (see Figure 6.1) are one

of the more economical shoring systems that can be installed for $15–$25 per SF

of shored face. The cost range is affected by whether steel or wood lagging is used,

whether it is cantilevered or braced, and whether the beams are drilled or driven, at

a minimum. Many other factors not mentioned here can affect cost and should be

looked at case by case.

The length of time many shoring systems are in place ranges from a few days to

several months to a couple years. Systems that are in use for 3 months or less benefit

from a short-term allowance. For short-term loading on shoring, allowable stress can

be increased by 33% except when:

Three months is exceeded.

Nearby construction equipment is causing vibration.

Excavations are adjacent to railroads.

Note: Strut design, which is covered later, is not offered this allowance.

6.2 MATERIALS 107

FIGURE 6.1 Soldier beam and lagging components.

Soil Arching Soil arching occurs when the supported soil is kept tight to the back

of the soldier pile (see Figure 6.2). This condition helps the lagging in that the lagging

does not experience the whole pressure of the soil. Engineers have concluded that the

soil pressure is only 60% effective; therefore, the maximum moment or pressure can

be multiplied by 0.60. Here is how the common maximum moment calculation is

affected by this reduction:

M = (0.6)wl2∕8, a 0.6 multiplier is used with lagging to account for soil arching.

The lagging must be strong enough to support the soil load and transfer this load

to each adjacent soldier beam. The two most common types of lagging are wood

(2 × 10′ to 6 × 12′) and steel plate (1′′ to 13∕4′′).Lagging calculations are somewhat simplified because the free-body diagram of a

single lagging board or steel plate is a true case of a simply supported beam, as shown

in Figure 6.3. The boards or plate rest behind the flange of the soldier beams. In rare

FIGURE 6.2 Soil arching between soldier beams.

FIGURE 6.3 Typical loading diagram for lagging.


cases where lagging cannot fit or for another reason cannot be installed behind theflange, other methods of attaching the lagging to the front face of the soldier beamare employed. For example, studs can be used on the face of the beam flange to hold asteel plate that captures the lagging. This would only be for wood lagging applicationsand is rare. In most cases, the maximum bending moment can be found by wl2∕8.

6.2.3 Lagging

Wood Wood lagging is used for temporary shoring walls and for permanentshoring walls. The primary differences between permanent and temporary use are(1) pressure-treated wood is normally used for permanent walls and (2) the boardthickness is greater for permanent walls due to the lower allowable stresses. Woodlagging is labor intensive, but the material costs are relatively inexpensive and thesystem is very versatile because the wood can be cut into any shape to fit aroundobstructions such as utilities and other shoring systems and stepped to go downslopes. Figure 6.4 also shows the versatility of wood for going down slopes.

Whenever wood is used for design, shear and bending are analyzed. However,because lagging boards are loaded in their weak direction (flat), shear is usually not aconcern. The boards loaded in this manner will always fail in bending before failing inshear. The cross-sectional area (shear) is the same either way, but the section modulus(bending) is very different between the strong and weak direction of application.

Wood Lagging Design A few common formulas (also mentioned in the first twochapters) are necessary for wood lagging design:

M = wL2∕8 (max bending moment for a simply supported beam)

FIGURE 6.4 Cantilevered soldier beam system.

6.2 MATERIALS 109

where w = uniform load

L = span from center of one soldier pile to another

Fb =MS

(max bending stress in a beam)

where M = max moment

S = beam section modulus

Fv = 1.5 V∕A (max shear stress, which is not necessary for lagging becausethey are always loaded in the weak direction)

where V = max shear from the FBD

A = cross-sectional area of beam

S = bd2∕6 (section modulus in the direction of loading shown as a 12′′ strip inFigure 6.5)

where b = base dimension

d = beam depth

Lagging boards always are assumed to be a 12′′ strip. This way, the units in calcu-lations shown above will always be in 1-ft increments. The same concept is used withsteel plate. In other words, it does not matter whether the lagging is a 2 × 8, 2 × 12,and so forth, it is always assumed to be a 12′′ strip. For this reason, the base (b)dimension in the section modulus formula is 12 in as shown in Figure 6.5. Therefore,the only unknown is the depth of the lagging. Figure 6.6 provides a good example ofhow a wood lagging system was used in a downtown location excavation. Let’s lookat some simple steps to determine the minimum depth of a wood lagging board.

Step 1: Determine the uniform load on one lagging board using w = (Pmax)(1 ft). For this example assume the soil pressure on the shoring system is600 psf

where Pmax = 600 psf soil loading

W = 600 psf × 1′ strip = 600 plf (since w in the moment

formula is pounds per linear foot)

L = 7 ft

Fb = 1800 psi for construction-grade Douglas fir−Larch

FIGURE 6.5 Typical lagging board loading.


FIGURE 6.6 Lagging system in place.

FIGURE 6.7 Lagging FBD.

Step 2: Draw an FBD (Figure 6.7) and calculate the maximum moment:

M = (600)(7′)2∕8 = 3675 ft-lb

Step 3: Set fb = M∕S and solve for d. The depth of the lagging is the only

unknown at this point.

Don’t forget to convert the feet to inches in the formula because the allowable

bending stress is in psi. The formula should look like this:

Fb = M∕(12′′ × d2∕6) (solved for d)

1800 psi = (3675 ft-lb)(12′′∕ft)∕ (12′′d2∕6)

d = 3.5 in

Conclusion: An S4S 4 × lagging board can be used because the actual depth is

3.5 in. Typically this would be a 4 × 12 so that the number of boards to be placed is

minimized for labor economy. The less number of boards to be handled, cut, placed,and removed, the less expensive the system. As an example, if the shoring wall was

15 ft high, the difference between 4 × 8′s and 4 × 12′s would be (25−16 boards)

9 boards.

6.2 MATERIALS 111

It should be noted that if the solution is not desirable, as with most temporary

structure design, the variable information can be modified in order to change the

results. For instance, in this problem we can change the type/specie of wood lagging,

which changes Fb to a higher or lower value (though 1800 psi is about as high as wecan go). We can also change the spacing of the soldier beams to lessen the bending

moment. We also did not take advantage of soil arching. If we did, we could have

reduced the 600 psf to 60% or to 360 psf. It should be determined if soil arching will

occur by consulting a soils engineer.

Steel Plate Steel plate lagging is more expensive to buy than wood lagging;

however, contractors often prefer it because of the savings in labor cost to install.

Like most decisions in temporary structures, the most economical system will be

chosen when considering materials, labor, and equipment needed. Sometimes the

more expensive material costs can be offset by lower labor rates. The down side to

steel plate lagging is its lack of flexibility to conform around obstructions and other

irregularities.

The same formulas as those used in wood lagging calculations are used. One big

difference is that for steel, Fv = V∕d × tw. This will not change the problem because

bending will always govern when the lagging is loaded in the weak direction (flat).

The other more applicable difference is that the Fb allowable bending stress for steelis so much more than for wood. The same 12′′ strip of steel is assumed as was the

12′′ strip with wood. This makes calculations simple and the units remain per foot.

Steel Plate Lagging Design Let’s design a steel plate lagging system with the

following parameters and using Figure 6.7 as a reference:

Pmax = 800 psf

W = 800 psf × 1′ strip = 800 plf

(since w in the moment formula is pounds per linear foot)

L = 8 ft

Fb = 21,600 psi for A36 steel (compact section 0.60Fy)

Step 1: Draw an FBD and calculate the maximum moment:

M = (800)(8′)2∕8 = 6400 ft-lb

Step 2: Set Fb = M∕S and solve for d. The depth of the lagging is the only

unknown at this point.

Don’t forget to convert the feet to inches in the formula because the allowable

bending stress is in psi. The formula should look like this:

21,600 psi =6400 ft-lb 12′′∕ft

12′′ × d2∕6d = 1.33 in


Conclusion: If steel plate is rolled in increments of1

8′′, then one could use 1

3

8′′

plate. If they come in1

4′′ increments, then we would need to use 1

1

2′′ plate because

1.33′′ is the minimum.

If soil arching was considered, we could have reduced the pressure to 480 psf

(0.60w).Steel plate lagging can be combined with wood lagging. Let’s say we have a

15-ft-deep excavation and we have 12 × 8 steel plates available. The plates would be

placed vertically, thus the top 3 ft would be unsupported. In this case, it makes sense

to place three to four wood lagging boards at the top of the shoring to get the system

to the original ground (OG) level. Figure 6.8 shows an example of this. The same

calculations would have to be performed. However, one should take advantage of the

fact that the wood lagging is at the top 3 ft of the excavation, which is where the soil

pressure is at its lowest value. Keeping in check with our minimum values, a reduced

pressure of say 400 psf may be acceptable at the top section.

Table 6.1 is from Caltrans Shoring Manual and shows recommended board thick-

nesses of Douglas fir with soil arching for different types of soils taken from the

Unified Soils Classification System. These values assume there is no surcharge load-

ing, only lateral soil loads. Surcharge loading is less common with these types of

systems, especially when they are supporting a bank or slope for a road widening

project where access to the top of the wall is impossible or impractical.

6.2.4 Soldier Beam Design

Once the lagging type and thickness is computed, the designer can consider the soldier

beam sizes. Soldier beams can be any W (wide flange) or HP (hinge point) shapes,

but HP shapes are usually considered first. HP shapes are available up to HP14 × 117,

FIGURE 6.8 Combined wood and steel plate lagging.

6.2 MATERIALS 113

TABLE 6.1 Recommended Lagging Board Thickness (without Surcharge)

Recommended Thickness of Wood Lagging When Soil Arching Will Be Developed (forlocations without surcharge loadings)

Soil Description

Classification Unified Depth

Recommended Thickness

of Lagging (rough cut)

for Clear Spans of

5′, 6′, 7′, 8′, 9′, 10′

Competent Soils

Silts or fine sand and silt

above water table; sands

and gravels

ML, SM – ML

GW, GP, GM,

GC, SW, SP, SM

0′ to 25′ 2′′, 3′′, 3′′, 3′′, 4′′, 4′′

(Medium dense to dense)

Clays (stiff to very stiff);

nonfissured clays, medium

consistency and

𝛾H∕C < 5.

CL, CH CL, CH 25′ to 60′ 3′′, 3′′, 3′′, 4′′, 4′′, 5′′

Difficult Soils

Sands and silty sands,

(loose); clayey sands

(medium dense to dense)

below water table

SW, SP, SM

Clays, heavily over-

consolidated fissured

SC 0′ to 25′ 3′′, 3′′, 3′′, 4′′, 4′′, 5′′

Cohesionless silt or fine

sand and silt below water

table

CL, CHML; SM

– ML

25′ to 60′ 3′′, 3′′, 4′′, 4′′, 5′′, 5′′

Potentially Dangerous Soils (appropriateness of lagging is questionable)

Soft clays 𝛾H∕C > 5; slightly

plastic silts below water

table; clayey sands (loose),

below water table

CL, CH ML SC 0′ to 15′

15′ to 25′

25′ to 35′

3′′, 3′′, 4′′, 5′′, 3′′, 4′′, 5′′,

6′′, 4′′, 5′′, 6′′

Source: Adapted and revised from the April 1976 Federal Highway Administration Report No. FHWA-

RD-130 and the CalTrans Shoring Manual.

which has a maximum section modulus in the strong direction of 172 in3. Beyond this

section modulus, wide flange (W shapes) need to then be considered. The method

of beam placement should also be considered. As mentioned earlier, beams can be

“drilled and dropped” and backfilled with slurry or driven with a vibratory or impact

hammer. The beams in Figure 6.9 were drilled and dropped. Technically, the place-

ment method would change the soldier beam design slightly because the beam width

is widened due to the slurry in the drilled hole; but in this text, the methods will result

in the same design.

The temporary structure student must first understand how the uniform load is

placed on the soldier beam. As we look at the load path that begins with the soil load,


FIGURE 6.9 Soldier beam and plate with dewatering system.

we notice that the soil loads the lagging and the lagging transfers that load to the

soldier beams. This means that each soldier beam has to support the load halfway toeach adjacent soldier beam. In other words, the soldier beam tributary width is the

same as the soldier beam spacing. Figure 6.10 shows this tributary width as it appliesto one soldier beam.

From this understanding, the designer must multiply the Pmax pressure by the trib-

utary width in order to continue with the soldier beam design. For example, if themaximum soil pressure is 800 psf and the soldier beam spacing is 6 ft on center, then

the uniform load on one soldier beam would be:

w = 800 psf × 6 ft = 4800 plf

The loading diagram would look like Figure 6.11.

Now that the uniform load on a single soldier beam is understood, the sizeof the beam can be determined. In order to size a beam, the maximum bending

moment needs to be calculated. Besides the uniform load, the designer would have to

FIGURE 6.10 Tributary width on a single soldier beam.

6.2 MATERIALS 115

FIGURE 6.11 Vertical loading on a single soldier beam.

determine the length of the beam and where it is supported. In Chapter 5 we studiedcantilevered beams and supported beams. Let’s look at the following examples.

Cantilevered Soldier Beams Cantilevered soldier beams must rely on a fixedportion of the beam supported by the passive soil pressure coming from the embed-ded portion of the beam below the subgrade (dredge line) of the excavation. For thispurpose, it will be assumed that the beam is embedded deep enough to develop fullfixity of the beam. This could be the 1.5–2.0 times the excavation depth mentionedin Chapter 5 and would provide a large enough FOS. Also, the beam fixity beginsat a point near the subgrade (SG), which can be relied on to offer a fixed connec-tion. An acceptable practice is to assume a point 3 ft below the dredge line, called thehinge point (HP). The 3-ft point is assumed so we can rely on a substantial passiveforce. This first 3 ft of soil from the SG to the HP is assumed to fail along a 45∘ fail-ure plane and cannot be relied upon for support. Figure 6.12 is a good example of acantilevered soldier beam system. The maximum moment for a cantilevered soldierbeam assuming proper fixity below subgrade is

M = wL2

2

It should be pointed out here that L includes the supported wall height plus theadditional 3 ft to the HP. In this example, if the excavation depth was 12 ft, thenL = 12′ + 3′ = 15 ft.

FIGURE 6.12 Soil loading on a cantilevered soldier beam.


Example 6.1 Soldier Beam CantileveredStep 1:Draw a proper loading diagram (see Figure 6.13) and determine themaximum

bending moment in the beam.

FIGURE 6.13 FBD of cantilevered soldier beam.

If the excavation depth is 12 ft from OG to the subgrade and the hinge point is 3 ft

below subgrade, then the full cantilever distance L is equal to 12′ + 3′ = 15 ft. Now,

let’s assume the soil pressure is the same as above with the 6-ft soldier beam spacing.

This resulted in a uniform soldier beam load of 4800 plf.

The resultant R = 4800 pfl × 12 ft = 57,600 lb or 57.6 k

The resultant location of a rectangular load is at themidpoint of the diagram height,

or 6 ft from the bottom of the excavation; but the 3 ft to the hinge point needs to be

added in order to represent the proper moment arm, which is from the HP to the

resultant.

Implementing two different cantilevered max moment formulas, the answers are

within 4% of each other. In most cases, the designer would be prudent to use the

higher, more conservative value of 540 ft-k. This would obviously produce the larger

soldier beam:

Mmax = P × L 57.6 k × 9 ft = 518.4 ft-k

Mmax = wL2∕2 4.8 klf × (15)2∕2 = 540 ft-k

Using a maximum moment of 540 ft-k, determine the minimum section modulus

of the soldier beam. It is preferable to find an HP beam first. However, if there is

not an HP beam with sufficient section modulus, then a W shape beam should be

considered:

Smin =MFb

where Smin = minimum section modulus

M = maximum moment in the beam

Fb = allowable bending stress of the steel

6.2 MATERIALS 117

If A36 steel is assumed and a compact section is desired, then Fb = 0.6Fy, or21.6 ksi:

Smin =540 ft-k (12′′∕ft)

21.6 ksi= 300 in3

The AISC manual contains most beams manufactured and their properties.

Appendix 1 should be used for beam selection. It is evident that there are no HP

shapes with a minimum section modulus of 300 in3. Therefore, the W shapes should

be considered. Since compact sections are desired for soldier beams, the designer

should begin with a 12′′ or 14′′ beam. For the most economical beam, the weight

per foot is a concern. The available W shapes are shown in Table 6.2.

TABLE 6.2 Available Beams Meeting Section Requirements

Nominal Depth (in) Beam Section Modulus (in3)

12 W12 × 230 321

14 W14 × 193 310

16 W16 × ? Not applicable

18 W18 × 158 310

Based on these results, the designer has the option of compactness vs. economy.

Since soldier beams are continuously supported by the earth and lagging, flange buck-

ling should not be an issue and compactness is not as important. As far as economy,

at steel prices between $0.55 and $0.75 per pound (at the time of this book’s publish-

ing), the lighter beams would save the project a great deal of beam costs. Let’s say

20 of these beams are required and they are 34 ft in length. The difference between

the W14 and the W18 using a steel price of $0.65/lb is

20 ea × 34 ft × (193 − 158 plf) × $0.65∕lb = $15,470

The same comparison can be calculated between the 12′′ and 14′′ beams:

20 ea × 34 ft × (230 − 193 plf) × $0.65∕lb = $16,354

This is a significant cost savings.

Supported Soldier Beam System Cantilevered soldier beams have their limi-

tations and where the depths exceed these limitations, supports must be added to the

system. Supports, in this case, are either struts or tiebacks.

Single Support/Tieback Figure 6.14 provides an example of a single supported

tieback system. Lets design another system using a single level tieback and the given

information in Example 6.2 that follows.


FIGURE 6.14 Single supported system.

Example 6.2 Supported Soldier BeamGiven information:

Soldier beams at 6 ft on center

Pmax = 1200 psf and the pressure diagram is triangular

Surcharge load of 100 psf as a rectangular load from OG to dredge line

Depth = 19 ft

Upper support located 6 ft from OG

Lower support is at HP, 3 ft below dredge line

Pmax = 1200 psf + 100 psf = 1300 psf

R1 (soil load) is located1

3up from bottom of dredge line

R2 (surcharge) is located halfway up from the bottom of the dredge line

Step 1: Calculate resultant, which is the area of the triangular load plus the 100 psf

rectangular surcharge load (see Figure 6.15).

R1 = 6-ft spacing × (1200 psf × 22 ft∕2) = 79.2 k located 22′∕3 frombottom = 7.3 ft.

R2 = 6-ft spacing × (100 × 22 ft) = 13.2 k located 22′∕2 from

bottom = 11 ft.

FIGURE 6.15 Total load from soil and surcharge.

6.2 MATERIALS 119

Step 2: Total the moments about the subgrade location (fixed point) on the right.

The drawing in Figure 6.16 has been turned 90∘ for simplicity. A shear

and moment diagram would be time consuming and could produce

errors.

Another method to determine the maximum moment in the beam that

is very conservative is to convert the soil and surcharge load to a uniform

load across the longest span of the beam and use the simply supported

beam moment formula (wl2∕8). In this case, the uniform load would be

1200 + 100 = 1300 psf over a 16-ft span (L). The load diagram would

look like Figure 6.17.

M = 6-ft beam spacing × (1300)(16′)2∕8 = 249,600 ft-lb

FIGURE 6.16 FBD of soldier beam.

FIGURE 6.17 Alternate FBD (simplified).

The beam size for this maximum moment would be

Smin =(249.6 ft-k)(12′′∕ft)

21.6 ksi= 138.7 in3

Table 6.3 lists some available beams and their beam section modulus in cubic

inches from Appendix 1.


TABLE 6.3 Available Beams

Nominal Depth (in) Beam Beam Section Modulus (in3)

12 W12 × 106 145

14 W14 × 90 143

14 HP14 × 102 150

16 W16 × 89 155

18 W18 × 76 146

Beam Software

Beam software can be used in this case as well. Figure 6.18 illustrates a computer-

generated diagram. Even though this is not an indeterminate structure, using software

such as this saves a lot of hand calculations. The surcharge load has been input as a

100-psf rectangular load, and the soil pressure is represented by a triangular load

ranging from 0 psf at the top to 1200 psf at the bottom. The upper hinged support is

a tieback 6 ft down from OG, and the lower support is fixed and at subgrade.

FIGURE 6.18 Computer-generated S and M diagram.

Max moment = 160.1 ft-k

Besides the program calculating the maximum moment in the beam, the program

also calculates other useful information. For example, the maximum shear value in

the beam is determined. In addition, the reactions are calculated, which in this case

determines the tieback load per 6 ft of wall. The shear and moment diagram shown

above indicates a 35.4-k tie load. If this was a strut, the load would be the strut force

per 6 ft of wall. So, if the tiebacks/struts were placed at every other soldier beam, the

load would be 2 × 35.4 k or 70.8 k.

There is an obvious difference between the two maximum moments. The con-

servative method produced a 249.6-ft-k moment and the beam program produced a

160-ft-k moment. These two methods led to two very different size beams. Econom-

ically speaking, this also results in a large beam cost difference.

6.2 MATERIALS 121

Soldier beams, depending on their size and the soil type can cantilever between

12 and 15 ft. This method can be very economical if the higher limits are reached.

For instance, a 15-ft cantilever with steel plates would be very inexpensive to install.

If the materials were owned by the installing company, then the costs of this system

could be very favorable.

Soldier Pile without Lagging

If the soil has enough cohesion, soldier piles can be spaced closer together and the

wood or steel lagging eliminated. The author is not recommending this method due to

the high cost of steel soldier pile; however, the goal here is to show different options in

construction. If a company owned these steel soldier pile and they were to be removed

at the conclusion of the project, then an argument (cost analysis) could be made to

justify such a system. Figure 6.19 shows this unique case where soldier beam spacing

was reduced to 3.5 ft on center to eliminate the need for lagging. Of course, the soil

type has to be cohesive enough to allow this type of system.

FIGURE 6.19 Soldier beams without lagging.

6.2.5 Tiebacks and Soil Nails

When excavation depths exceed the option to cantilever (up to approximately 15-ftdepth) or to strut to the opposite side of the excavation, the need for “tying-back” isusually one of the only options. When there is no obstruction underground such asutilities, a swimming pool, roadways, a building foundation, high groundwater, andthe like, tiebacks are fairly versatile.

The sketch in Figure 6.20 shows the three main components of a tieback system.The soil or rock is what is drilled through to create the hole for the tieback. The grout


FIGURE 6.20 Tieback isometric.

FIGURE 6.21 Two-pair strand tieback system.

joins the soil or rock to the steel tension members (strand or rod). Finally, the tension

members, which are rods or strand, join the grout to the structural framework of the

shoring, such as walers or soldier beams.

Figure 6.21 provides an example of a two-pair strand system. The materials that

make up these systems are usually high-strength steel ranging in allowable stresses

between 60 and 120 ksi.

Determining Tieback Loads from Soil Pressure Diagrams Pressure dia-

grams are essential for designing beam components, but they have other benefits as

well. The reactions that are calculated on a pressure diagram typically represent the

support of the beam in question. Since the next step of the design is to size the support

system or apply the reaction load to the next member, these values give you the infor-

mation needed for this step. For instance, the information from Example 6.2 can be

used in the next step. If we look a little closer, we notice the reactions at the left and

right are 35.4 and 57 k, respectively. The values could represent either the forces from

this beam to its supporting beam or the forces required in a tieback or strut system

6.2 MATERIALS 123

if this is the final portion of the design. Shoring loads always terminate at a tieback

or strut.

Once the system load requirements are known, the tieback can be designed. The

factors that contribute to the strength of a tieback are the length (L), the shear strengthof the soil or rock (psi or psf), the diameter of the drilled hole, and the strength of the

steel tension members. In addition, a factor of safety should be included between 1.25

and 2.5 : 1.0. A good tieback design would closely match the strength of the tension

member to the strength of the grout vs. soil/rock matrix. Also, it is assumed that the

grout material matches or exceeds the strength of the soil/rock.

The strength of a tieback can be determined by combining the factors mentioned

above into a formula. The formula, shown below, basically determines the contact

surface area of the earth and the grout. First, the circumference of this contact zone

multiplied by the length of the hole gives us the area of this contact surface in square

feet. Next, the shear strength in psf is multiplied in order to determine the allowable

load of the system. Finally, an FOS is divided into this load in order to obtain the safeworking load (SWL) of the anchor.

Pall = L × D × t × 𝜋 ∕ FOS

where Pall = SWL of the anchor

L = length of the anchor

D = Diameter of the drilled hole

t = shear strength of the soil or rock

𝜋 = 3.1416

and FOS is between 1.25 and 2.5.

Most of the components of this formula are self-explanatory, except for shear

strength of soil or rock. The values used for shear strength have been studied in great

length. The shear strength of the soil is a main component of the tieback strength,

which is shown in section in Figure 6.22. The failure plane is shown in this figure

and where the bonding zone of the tieback is located.

FIGURE 6.22 Soil anchor embedded behind failure plane.


FIGURE 6.23 Soldier beam shoring with tiebacks.

Shear Strength of Soil and Rock According to theories from Mohr (1900) and

Coulomb (1776), material fails from a combination of shear and normal stresses.

Depending on the material type (cohesion and angle of internal friction values

varying), the shear strength can fluctuate due to these stresses. Like everything else

we have seen so far, when soil properties are different—whether granular, cohesive,

or silty—all will produce different strength characteristics. Similarly, very soft to

very hard soils will produce a range of different strengths the designer can rely on

to estimate the strength of a soil/rock anchor or soil nail. Figure 6.23 shows two

strand tiebacks used with a cantilevered soldier beam and lagging system for a road

widening project.

For literally hundreds of years, engineers and scientists have been developing the-

ories about soil strengths. The intention of this text is to minimize confusion and

illustrate the basic concepts behind the subject. In this case, we need to understand

what kind of shear strength one can expect from different soils and how one deter-

mines allowable loads on a system relying on these shear strengths. Table 6.4 shows

TABLE 6.4 Sample Material Strengths

Shear Strength Approximate psf Approximate tons/SF

Very hard rock >5200 >2.6

Rock 3000–5200 1.50–2.6

Hard soil 3140 1.57

Stiff soil 1560–3140 0.78–1.57

Firm soil 840–1560 0.42–0.78

Soft soil 420–840 0.21–0.42

Very soft soil <420 <0.21

6.2 MATERIALS 125

a range of material strengths. This table is just a guide; actual strengths should be

derived from field and laboratory testing.

Using the theory of soil nail design above, the following examples will work

through the process of either determining soil anchor capacity or determining the

required length of a soil anchor.

Example 6.3 Determine the allowable capacity of an anchor with the following

characteristics:

Pall = ?

Shear strength of the soil is 1800 psf

Length is 30 ft

Diameter of hole is 4′′

FOS = 1.5

Pall = (30 ft)(4′′∕12′′∕ft)𝜋(1800 psf)∕1.5Pall = 37,700 lb (37.7 k)

Example 6.4 Determine the necessary length of a rock anchor requiring 10,000-lb

capacity:

L = ?

Required P = 10,000 lb

Shear strength of the soil is 4000 psf

Diameter of hole is 2′′

FOS = 1.25

10,000 lb = (L)(2′′∕12′′∕ft)𝜋(4000 pfs)∕1.25Solving for Lmin

L = (10,000 lb)(1.25)∕(2′′∕12′′∕ft)𝜋(4000 pfs)Lmin = 5.97 ft long

In Chapter 5, the subject of slip planes was discussed, and it was mentioned that

tieback lengths should be measured from behind the slip plane so that the unstable

wedge of soil does not influence the failure of the tieback. It is also common practice

in temporary and permanent design to sleeve the hole that passes through the fail-

ure wedge to avoid aiding this wedge into more failure. In the examples above, the

distance L is the anchored length of the soil anchor. This concept is well illustrated

in Figures 6.22 and 6.24. If the thickness of the failure wedge is added, then the

drilled hole would be L′ = L+ “wedge thickness.” Figure 6.24 shows these various

dimensional properties of the bonded and unbonded lengths.


FIGURE 6.24 Bonded and unbonded zones of a tieback.

CASE STUDY: SOLDIER BEAM AND PLATE

While BART was being extended from Fremont to San Jose, California,

there were many challenges to the excavations along some of the narrow

right-of-ways. Several different support of excavations were employed. Along

one 1000-linear foot long stretch, a soldier beam and plate system was put in

place.

Equipment used is shown in Figure 6.CS1.

FIGURE 6.CS1 Soldier beam and plate system.

6.2 MATERIALS 127

Plate installation is shown in Figure 6.CS2.

FIGURE 6.CS2 Plate installation.

FIGURE 6.25 Deep soil mixing equipment Presidio Project, San Francisco.


Other Support of Excavation Methods This chapter has discussed one of

the more common types of shoring systems when excavation depths do not exceed

30–40 ft. Figures 6.25–6.29 show other support systems for projects requiring exca-

vation. These systems were not discussed in this chapter; however, they are frequently

used in the industry. The next chapter will review sheet pile and walers in more detail.

FIGURE 6.26 Multilevel walers and struts supporting soldier beam and deep-soil mix wall.

FIGURE 6.27 Deep soil mixingwith soldier beams, Owner, Valley Transit Authority (VTA).

6.2 MATERIALS 129

FIGURE 6.28 Valley Transit Authority.

FIGURE 6.29 Soldier beam and lagging wall with tieback and no waler, Enloe Hospital,

Chico, California.

CHAPTER 7

SHEET PILING AND STRUTTING

7.1 SHEET PILING BASICS

7.1.1 Materials

In this chapter, the materials discussed will follow the same allowable stress valuesas Chapter 6. Sheet pile systems are predominantly steel systems. The sheet pile areeither hot or cold rolled steel, walers are WF beams or double steel channels, andstruts are steel pipe, wide flange (W), or HP beams. As mentioned in Chapter 2,there are different types of steel for temporary and permanent construction. The gradeof steel can range from 36,000 psi to over 80,000 psi. As in the previous chapters,A36 steel will be the most widely used steel type.

7.1.2 System Description and Unit of Measure

When describing sheet pile systems, the load path can help explain the componentsin the order in which the load travels. The soil pressure is applied to the sheet pileitself. In this case, as in most cases, the sheet pile acts as the sheeting material. It isthe first contact between the soil and the system itself. Typically, the sheet piles areinstalled before the excavation begins. Therefore, the sheeting does not encounter thepressure until the excavation is deep enough to develop soil pressure at a particulardepth. At this point in time, the sheet piling either has to be designed to support thesoil pressure or the excavation stops in order to install a waler/strut system. Figure 7.1illustrates the sequence of a typical excavation with one level of bracing.

Hot rolled sheet piling, shown in Figure 7.2A, is the more desired sheet pile for itsstrength and interlock tightness. As already mentioned, the cost for hot rolled sheetsis slightly higher but that comes with better quality.

130

7.1 SHEET PILING BASICS 131

FIGURE 7.1 Excavation sequence.

FIGURE 7.2A Hot rolled sheet pile.

Sheet piling comes in two basic forms: hot rolled and cold rolled. Hot rolled sheets

are formedwhen the steel is moltenmetal (liquid). The interlocks are shapedwhile the

steel is formable. This makes the interlock tight, which means less water or soil can

penetrate the interlock joint. The cold rolled sheets are bent from flat plate when the

steel is cold. The interlocks are bent as shown in Figure 7.2Bwith machinery to create

FIGURE 7.2B Cold rolled sheet pile.

132 SHEET PILING AND STRUTTING

the interlock. These sheets are less expensive; however, they are not as watertight asthe hot rolled sheets.

Sheet piles are quantified by the number of pairs to be driven and by the totalpounds of steel to be purchased or rented. The contractor has to install each pairand remove each pair by the method chosen and described in the next section of thischapter. In addition, sheet pile are either purchased for $0.60–$0.90 per pound orrented for a monthly rate per pound. When estimating this type of work, the sheetpile liquidation cost should be added to the price of the sheet piling. Liquidationis the amount of sheet pile that gets damaged and cut off during installation andremoval procedures. For example, a contractor rents 40-ft-long sheet pile. Then afterthe project is completed and the sheets are extracted, there is typically about a 2-ftsection of sheet that is damaged due to the installation and extraction process. A dam-aged sheet pile is often unusable due to a damaged interlock. The 2 ft that becomesscrap metal is charged the liquidation price. It is seldom economical or feasible torepair the damaged 2 ft of the sheet and it is difficult to justify the labor cost.

Walers Walers (sometimes referred to as wales) are horizontal steel beams thatsupport the sheet piling when the depth becomes too much for a cantilevered system.Walers can be any steel sections such as wide flange, HP shapes, or double channelsat a minimum. If a strutting system is going to be used, then wide flange and HPshapes are common. If a tieback system (mentioned in Chapter 6) is used, then doublechannels will allow a space for the tieback rod or strand to pass through and load thewaler concentrically. Figure 7.3 shows a double channel waler sytem.

Strutting Strutting accepts the load from the waler and transfers this load to thewaler on the other side of the excavation or cofferdam. Struts are loaded axially,

FIGURE 7.3 Double channel waler system before excavation.


FIGURE 7.4 Sheet pile, waler, and strut system.

so they need to have substantial buckling resistance. Pipes are very popular as struts

because they have the same radius of gyration 360∘ around. Square tubing offers thesame symmetry with an increased cost. In comparison, a beam, which has a strong

rx and a weak ry, always has to be checked for the weak axis or braced along the

weak axis.

Figure 7.4 shows a three-level waler and strut system. The waler is a W24 wide-

flange beam at 12-ft centers and the struts are 18-in-diameter pipe at 16-ft centers.

These levels were placed as the excavation proceeded downward. Struts always pose

an excavation production hindrance because the excavator has to work between them

and take care not to make contact with them, which typically slows down produc-

tion. Tiebacks (also discussed in Chapter 6) can be used in lieu of strutting. However,

tiebacks can be much more expensive, and they disrupt the production of the excava-

tion more because they take longer to install than struts. The upside to using tiebacks

is that they leave the contractor more room to work within the excavation.

7.1.3 Driving Equipment

Although this is a design textbook, it is sometimes important to briefly discuss

installation methods. Three methods of sheet pile driving are briefly covered here:vibratory, hydraulic impact, and diesel hammer impact.

Vibratory Z-shaped sheet piles are generally driven as pairs. This means that two

Z sections are threaded together and driven as a pair (at the same time). The hammer

secures and impacts the center of the pair, just above the interlock. The traditional

vibratory hammer is the most common method of driving sheet piling, as long as the

soil type is favorable. With fairly soft to medium compacted soils, sheet piles can


FIGURE 7.5 Pile driving operation using vibratory hammer.

be installed with good success. Figure 7.5 shows a pile driving operation using anApe 200 vibratory hammer. There are also fewer chances of obstructions with soil ofthis type.

Hydraulic Hammer When the soil type is harder than a vibratory hammer canpenetrate, the use of impact hammers becomes necessary. This method is only usedwhen a vibratory hammer is not practical. With this method, more time is spent inthe lining, driving, and cutting or “fresh heading” the sheets when damage occurs.Fresh heading means the top of the pile is damaged; therefore, the driving has to stop,the hammer is removed, and the pile driving crew cuts the bent part of the sheet pile inorder to start with a flat, uncrimped top. This process is costly as it interrupts produc-tion of driving the sheets. However, if the operation continued to drive the damagedpile, even more time would be lost in production because the energy between thehammer and the pile would be lost.

Hydraulic hammers can be single and double action. This means the hydraulicram either pushes one direction (down) or both directions (up and down). The impactthe hydraulic hammer provides is not as violent as the diesel hammer mentionednext. The force that this type of hammer provides can also be regulated in order tominimize the damage to the top of the sheet pile or slow the hammer down in sensitiveconditions. The double-acting hydraulic hammer has more strokes per minute, soeven though it does not impact with as much force, part of this can be made up withmore blows to the sheet pile per minute. The lower impact also lowers the chance ofcrimping the top of the sheet pile. The hydraulic hammer does require a hydraulicpower pack to operate the system’s hydraulic fluids. This becomes an extra piece ofequipment to the operation.

Diesel and Steam Hammer The diesel and steam hammers are the lastmethod that a pile driving crew would prefer to use. The diesel hammer can also


FIGURE 7.6 Sheet pile driving with an impact hammer.

be double acting. The time to loft the pile, place the hammer on the pile, and drivethe pile takes much longer than the vibratory hammer and a bit more time than thehydraulic hammer. Figure 7.6 shows a sheet pile driving operation using a DelmagD36 diesel hammer. With the diesel hammer, fresh heading is very common and isprobably necessary one to three times per sheet pair, depending on the soil type andexperience of the crew. The impact force that the hammer supplies has a great dealof energy (up to 90,000 ft-lb); and if the hammer is not perfectly plumb, the top ofthe sheet pile will crimp, thus require fresh heading. Diesel and steam hammers donot require a hydraulic power pack to operate the hydraulics.

Templates The quality of the installation of sheet piling ranges from not veryimportant to very critical. If the excavation is designed to be much larger than theroom needed to work on the permanent structure, then the location and vertical align-ment of the sheeting is not that critical. However, if the sheeting is designed as apermanent structure, or there is little room between the permanent structure and thesheeting, then quality control can be very critical. When the latter is the case, a sheetpile template can be used, as shown in Figure 7.7. A template can be made manydifferent ways. They can be multiple tiers tall or a single beam lying on the groundsurface. External walers (which will be covered later) can be used as a single-tiertemplate because they rest on the ground level and can be used as a template as wellas function as a waler. This is similar to the method shown previously in Figure 7.5.

The idea behind the template is that it is surveyed in and plumbed using preciseinstrumentation and then used to guide the sheets into their final location. The tem-plate needs to be designed strong enough that if the sheet pile bumped it, it would notlose line or plumb. This does not mean that it has to be designed by a certified engi-neer, but is should be checked by an experienced superintendent, project manager,or project engineer.


FIGURE 7.7 Two-tier driving template.

Driving Sheets in Hard Soils Hard soils can be very difficult to drive sheet pile

into unless the contractor takes one or some of the following measures:

Incorporate a heavier sheet pile section.

Jet the tip of the sheet pile with high pressure.

Penetrate the sheet pile line with a spud beam.

Use an impact hammer to drive the sheets.

Predrill the sheet pile line with a full flight auger, as depicted in Figure 7.8.

Preexcavate the inside of the cofferdam to reduce skin friction.

All of the above-mentioned methods are costly. The decision to use sheet piles in

the first place was derived from other factors, and lowest cost is usually not one of

the reasons. In many cases, sheet piling is the only option, so the goal is to make the

system as inexpensive as possible.

Two types of designs will be illustrated in this chapter: the cantilevered sheet pile

system and the braced sheet pile system. The soil loading has been previously dis-

cussed so the soil loading values used for the following examples will be given by

the author, and it will be assumed that they were calculated using the methods in

Chapter 5.

Cantilevered Sheet Piles The cantilevered method for sheet piling simply

means that there is no need for a waler/strut brace, and the sheets can support the

pressure on the cantilever without failing in bending. A rule of thumb for excavation

depths using the cantilevered method is between 10 and 18 ft, depending on the

soil type and sheet pile size. A cantilevered system relies heavily on the amount

of sheet embedded below the dredge line (bottom of excavation) and the soil type


FIGURE 7.8 Predrilling a sheet pile line in the Feather River.

FIGURE 7.9 FBD of a cantilevered sheet pile.

as the passive pressure on the toe of the sheet pile. Because of this reliance, the

passive shear strength of the soil is crucial as it pushes against the back of the sheet.

Once embedded properly, the sheet pile is “fixed” at the hinge point, which, in this

book, is 3 ft below the dredge line. From this point, the soil load is rectangular or

triangular, depending on the soil type, and applies this uniform pressure to the back

of the sheets similar to Figure 7.9.

Sheet Pile Design with a Cantilever Sheet piles are like other sheeting materials

in temporary structures because they are analyzed by a 12-in strip. The sheet pile

tables that will be used for sheet pile design are in Appendix 5. These tables consist

of lists of AZ and PZ sheet piling. The AZ and PZ designations are specific to the


manufacturers, and these are just two of several manufacturers. The tables provided

by these manufacturers basically have the same information categories, except the

values are going to be different per sheet type. The following is the most common

information needed about sheet piles:

The section name (PZ 22, AZ 17, etc.)

The sheet dimensions (width, depth, and thickness)

The weight per square foot or linear pile (psf, plf)

The section modulus (in3)

Sometimes the section name matches another property of the sheet pile. For

instance, the PZ sheet sections are PZ 22, PZ 27, and so forth. The 22 and the

27 represent the sheet’s respective weight per square foot. This makes calculating

weight very quick for the designer and estimator.

The first sheet pile design example will be a cantilevered system.

Example 7.1 Design a sheet pile with a uniform load of 1150 psf and cantilevered

as shown in Figure 7.9. Add the excavation depth to the 3-ft hinge point to achieve

the maximum cantilever.

Maximum moment at fixed end of sheet pile that can’t rotate is M = wL2∕3:

M = (1150 ft)(14′ + 3′)2

3= 110,783 ft-lb

Smin =Mmax

Fb

Smin =(110,783 ft-lb)(12′′∕ft)

21,600 psi= 61.55 in3

According to Appendix 5, the AZ 37-700 has a section modulus of 68.9 in3.

This sheet would be acceptable, but at 36.33 psf, the steel costs may be higher than

desired. On the other hand, if the alternative is adding a waler and strut system, an

estimate would have to be done comparing the larger sheet costs to the lower sheet

costs with the added waler and strut system.

Braced Sheet Piles Abraced system relies on the sheet strength but also is depen-

dent on most of the forces transferring to the walers. In this type of system, the toe or

embedment of the sheet pile is not as deep. This depth is from the subgrade/dredge

line to the tip of the sheet. Figure 7.10 shows a sheet pile operation prior to the exca-

vation being complete. The waler has been installed and the tiebacks are in progress.

Once the tiebacks are stressed, the excavation can continue.

Braced Sheet Pile Design The maximum moment of the sheet pile with two or

more spans can be estimated using M = wL2∕10. This is the basic bending moment

for beams with multiple supports. If there are only two supports, then M = wL2∕8.


FIGURE 7.10 Double channel waler system with tiebacks.

Example 7.2 Design a sheet pile with a uniform load of 1500 psf and supported as

shown in Figure 7.11. Add the excavation depth to the 3-ft hinge point to achieve the

maximum space between walers. Use this distance for L if it is the longest unsup-

ported length. Since 12 ft + 3 ft is greater than any of the other spans of sheet piling,

then L = 15 ft.

FIGURE 7.11 Supported sheet pile system.

The first step is to determine the maximum moment using the more-than-two sup-

port moment formula:

M = (1500 ft)(12′ + 3′)2

10= 33,750 ft-lb

Smin =Mmax

Fb


Fb = 0.6Fy Fb = 21,600 psi

Smin =(33,750 ft-lb)(12′′∕ft)

21,600 psi= 18.75 in3

According to Appendix 5, the AZ 12 has a section modulus of 22.3 in3.

The PZ 27 has a section modulus of 30.2 in3 (the PZ 22 is slightly under the

22.3 in3 with 18.1 in3).

It becomes apparent that supported systems can use lighter sheet piling even when

the loading is greater. This is mostly due to the reduced bending moment. This same

example with only a top waler and the lower hinge point would produce a larger bend-

ing moment; but without doing a calculation, it is difficult to know by how much. If

the excavation was the same depth, this would pose a large challenge because Lwouldbecome 36′ + 3′ = 39 ft. However, the bending moment would then be M = wL2∕8.

Rework the previous example with the following new information:

M =(1500 plf)(39 ft)2

8= 285,188 ft-lb

Smin =(285,188 ft-lb)(12′′∕ft)

21,600 psi= 158.44 in3

Appendix 5 does not have a sheet pile with such a large section modulus. There

are several choices at this point:

Redesign with a smaller sheet pile combined with a steel pipe pile also know as a

combi-wall.

Lower the upper waler to reduce the bending moment on the sheet.

Add at least one row of struts and walers.

It should be noted that the methods of determining the maximum moment in the

examples above are somewhat conservative. If a more accurate moment is required,

then the designer should:

Employ engineering methods for indeterminate structures.

Employ a beam software program as mentioned earlier in this text.

Walers Walers are the structural members that transfer the loading from the sheet

pile to the struts or tiebacks. The waler design is dependent on the uniform load,

which is determined by its spacing and the Pmax load from the soil. The further the

walers are from one another, the larger the linear load. The greater the soil pressure,

the larger the linear load as well. Figure 7.12 shows the amount of load each waler

is supporting. The linear load on the L2 waler is determined by the soil load times

the uniform width of the L2 waler. The design on the waler is also dependent on the

spacing of the struts because the struts are the supports of the walers and dictate L.A waler can be a simply supported beam if it only has two supports, one at each end.


FIGURE 7.12 FBD multilevel support system.

A waler can also be a continuous beam with multiple supports (struts). Whichever is

the case, the linear load will have to be determined.

Since walers make continuous contact with the sheet pile, lateral buckling is

usually not a concern. The two are usually welded in some fashion to temporarily

hang the waler and keep it from falling into the cofferdam before the soil pressure

compresses it with the struts.

In the sketch shown in Figure 7.12, assume the distance between the L1 waler andthe L2 waler is 12 ft and the distance between the L2 waler and the HP at the bottom

is 16 ft.

The tributary width on L2 is half the distance to L1 plus half the distance to the HP:

L = 1

2(12′) + 1

2(16′) = 14 ft

If the average uniform load from the soil pressure is 1200 psf, the linear load can

be determined by

W = 1200 psf × 14 ft = 16,800 plf

Once the linear load on a waler is computed, the first part of the waler design

is complete. The second part is determining the support condition of the waler. As

mentioned above, it can be simply supported if the shoring wall is not very long, or it

can be a continuous beam with multiple supports, which in this case would be struts

or ties (Figure 7.13). As we have learned, with the same uniform load (W) and the

same unsupported lengths (L) the maximum moment of [7.3(a)] a simply supported

FIGURE 7.13 Support conditions for a waler.


beam and [7.13(b)] a continuous beam with multiple supports is different, and thecontinuous beam will always be less if the conditions described are true.

Using the linear load calculated above and L = 16 ft, determine the bendingmoment of walers in Figure 7.13(a) and 7.13(b).

[7.13(a)] M = wL2

8

M = 16,800(18)2

8= 680,400 ft-lb

[7.13(b)] M = wL2

10

M = 16,800(18)2

10= 544,320 ft-lb

Example 7.3 Using the case in Figure 7.13(b), determine the minimum size L2waler (16,800 plf) that is required for bending. Assume the waler is attached to the

sheet pile continuously and lateral buckling is not an issue.

Smin =Mmax

Fb

Fb = 0.6Fy Fb = 21,600 psi

Smin =(544,320 ft-lb)(12′′∕ft)

21,600 psi= 302.4 in3

Refer to Appendix 1 and select three beams that have a section modulus equal to

or greater than 302.4 in3. Some beams that may work here are W18 × 158, W21 ×147, W24 × 131 and W30 × 129.

Internal Walers Internal walers are the most common type of waler in cofferdams,whether they are on dry land or in a marine environment. The sheet piles are loadedfrom the outside, inward. So it makes sense to allow the structural components topush against each other. The size of the waler determines the spacing of the struts thatsupport the waler. Another way to look at this is to say the strut spacing determinesthe waler size. Like most decisions in temporary structures, the spacing and sizesof materials makes up a large portion of the total cost of the system. The larger themembers, the less number of pieces required. On the other hand, the smaller membershave to be spaced closer together, thus requiring more pieces to complete the system.The final decision can’t be made until a complete quantity takeoff is performed and acost analysis done for each option including materials, labor, and equipment. Then,at this time, an educated decision can be made.

Internal walers can be used as driving templates and then converted to a waler oncethe sheets are in place. Figure 7.14 shows a river cofferdam. The additional pile (pinpile) inside the cofferdam are used to temporarily support the template (waler) duringthe sheet pile driving process. Once the sheets were driven, the template is attachedto the sheet pile and the pin piles are removed. The template now becomes the walerand the excavation can begin.


FIGURE 7.14 Internal waler in a river cofferdam.

The downside to internal walers is that the waler takes up more room inside thecofferdam. The cofferdam usually needs to be larger because the waler must clear thepermanent construction. A larger cofferdammeans there will be more excavation andbackfill quantity and also means at least four more pairs of sheet pile (one for eachside = 3–4 ft) have to be driven and removed.

External Walers External walers are used when there is only support needed at thetop of the sheet pile and the sheet pile can handle the bending stress from the top(OG) to the hinge point, 3 ft below the dredge line (subgrade). Figures 7.15 and 7.16are two examples of external waler systems.

The clamps that attached the top of the sheet to the waler can be a bridge style Cclamps shown in Figure 7.15, a section of steel plate cut into a C shape (pac man) asshown in Figure 7.16, or even a coil rod installed into a burned or drilled hole throughthe sheets and walers. The first two options cause less damage to the steel beams andsheet pile. Whenever possible in temporary construction, damage to the steel shouldbe minimized or eliminated all together. When steel is cut unnecessarily, one or moreof the following occur:

The steel has to be repaired if possible.

The member can’t be used in the future to its maximum design potential.

The member is shorter than its original length.

The contractor has to pay liquidation costs.

For the same reason an internal waler causes a larger cofferdam, the external walersystem allows a smaller cofferdamwhen there is only one top level waler. The smallercofferdam reduces the earthwork and sheet pile quantities, thus making the excava-tion, as a whole, more economical. Also, the excavation crew does not have to work


FIGURE 7.15 External waler system using bridge style C clamps.

FIGURE 7.16 External waler using plate clamps.

around the internal waler and risk damaging or compromising its strength. In addi-

tion, while the permanent structure is being built, the trades have clear access with

equipment and ladders. Finally, if there is a concrete foundation to be placed inside

the excavation, the concrete can be poured against the sheet pile, thus eliminating

additional internal support and the cost of formwork and back filling within a small

space (between the footing and sheet pile). If needed, a strut can still be installed with


FIGURE 7.17 Strut supporting walers.

external walers as long as the struts make contact with the walers and not just the sheetpile. Figures 7.15 and 7.16 show how difficult it would be to add an internal strut.

Struts So far in this text we have looked at some axial loads on columns and struts.Figure 7.17 shows a very short strut used in a sheet pile cofferdam system. We know

that the design of a strut is determined by:

The axial load

The unsupported length of the strut

The weakest radius of gyration

The end connections

With all of the information above, a strut can be designed for any system. There

are cases where practicality tells the designer that the strut is extremely long andeither can’t carry the axial load without buckling or needs to be supported for its ownweight and weak direction axis. These problems can be overcome with proper design

considerations.In Chapter 2, column and strut design was introduced. In this section, the appli-

cation learned earlier will be applied to a practical case. The first challenge whendesigning struts is to determine the axial compressive load on the strut. This can be

accomplished with one of three methods:

Using the linear waler load per foot

Calculating the pressure within the tributary area of one strut (see Figure 7.18)

Adding the reactions of the free-body diagram from each side of the strut


Tributary Width

1500 psf

Tributary Height

FIGURE 7.18 Tributary area of a single strut.

Example 7.4 Using the load in the previous waler example, determine the load on

a strut if they are spaced at 18 ft on center.

No matter which method is used, the same result should occur.

Method 1: Using the linear waler load from the previous waler example, multiply

this linear load times the strut spacing:

P = 16,800 plf × 18 ft on center (OC) = 302,400 lb

Method 2: Using the design load of 1200 psf, multiply this times the tributary

area on one strut. P = Pmax × Trib. width (strut spacing) × Trib. height

(waler spacing):

P = 1200 psf × 14 ft × 18 ft = 302,400 lb

Method 3: Add the left reaction to the right reaction from the free-body diagram:

151,200 lb + 151,200 lb = 302,400 lb

These three methods should match; if they do not, a mistake has been made. Now

that the axial load has been determined, the unsupported length of the strut is needed.

This is usually achieved by calculating the cofferdam size and subtracting the waler

depth on each side. In other words, if the cofferdam is 30 ft across, sheet pile to sheet

pile, and a W24 is used for the waler system, then the strut length would be approxi-

mately 30′ − 2′ − 2′ = 26 ft long. This length will be used for the rest of this example.

Figure 7.19 shows struts used in a floodplain cofferdam.

To determine the size strut required, the trial-and-error method can be used. This

means that the designer will choose a steel member, either a beam or a pipe and check

to see if it works. As in Chapter 2, Appendix 4 will be used. This table has precalcu-

lated Fb (ksi) values using a factor of safety and A36 steel with E = 29,000 ksi. The

first trial and error will be for an HP beam and the second will be with steel pipe.

These values can be found in Appendixes 2 and 3. However, if the pipe used is larger

than those shown in Appendix 2, the value will have to be calculated using common

formulas for r and A. The following steps will be used:

1. Assume ry = 3.0′′, weak direction radius of gyration. This is an average of a

range of beams that are commonly used as struts or columns.


FIGURE 7.19 Struts used in a floodplain cofferdam.

2. Using ry = 3′′ and k = 1.0, calculate the slenderness ratio (SR) of the strut

(SR = kl∕r).3. Using Appendix 4 for allowable buckling stress, determine themaximum buck-

ling stress of the H pile beam.

4. Using f = P∕A, calculate the required (A)rea of the cross section.

5. Find an H pile section that is close to this required (A)rea in Appendix 3.

6. Recalculate the new SR with the actual ry.

7. Go back to Appendix 4 for the new, actual Fb (ksi).

8. Using f = P∕A, look up the area of the new strut in Appendix 3, and calculate

P allowable.

9. Compare Pall to Pactual.

If Pall < Pact, repeat the last five steps (5–9). When Pall > Pact, the strut works.This method will always find a strut that is adequate for the load, but it is not

guaranteed to find the best or most economical strut. In step 5, when the designer is

looking for a strut to match the minimum area required, there are several beams from

which to choose. Since the area is used just to get the designer close, there are cases

when a slightly lower area will produce an adequate strut and be more economical

because a lower area will always be lighter per linear foot. Since steel is purchased

by the pound, economy means less steel weight.

Example 7.5 Using the method described above, find an HP and a pipe strut that

work to support 303 k with Le = 26′.


H Pile Strut

Assume ry = 3′′, k = 1.0

SR = [(1.0)(26′ × 12′′∕ft)]∕3.0′′ SR = 104

Appendix 4, Fb = 12.47 ksi

Using f = Pact∕A and Pact = 302 k, solve for A; A = 303 k∕12.47 ksi = 24.3 in2.

Select an HP shape in Appendix 3 with an area close to 24.3 in2, say HP 13 ×87 (25.5 in2)

HP 13 × 87 has an ry = 3.13′′ New SR = [(1.0)(26′ × 12′′∕ft)]∕3.13′′ SR =100 Fb for a 26-ft-long HP 13 × 87 is 12.98 ksi

Using an HP 13 × 87, A = 25.5 in2, Pall = 12.98 ksi × 25.5 in2 = 331 k > 303 k

(OK)

Steel Pipe Strut

When the designer is using steel pipe, the pipe properties have to be available.

Sometimes this is not the case, so the properties have to be achieved using statics.

The following is a review of the area (A) and radius of gyration (r) of a cylinder,

where Do (OD) is the outside diameter and Di (ID) is the inside diameter.

One can determine the area and radius of gyration of a cylinder by using the

following formulas:

A = 1

4𝜋 (D2

o − D2i )

rx = ry =1

4

√D2o + D2

i

Let’s look at an 18′′ × 1

4′′ wall pipe strut as shown in Figure 7.20 to support the

long sides of this cofferdam. If a large pipe is designated as being 18′′ in diameter,

the 18′′ dimension is the outside diameter. Therefore, the inside diameter is 18′′ −(2 × 1

4′′)= 17.5′′. Using this information, the area and radius of gyration can be

determined:

A = 1

4𝜋 [(18)2 − (17.5)2] = 13.94 in2

rx = ry =1

4

√(18)2 + (17.5)2 = 6.28 in

Trial and error is more difficult if the designer does not have a list of the pipe

properties. For example, with pipe, a radius of gyration assumption of 3.0 would be

very low. As you can see, the pipe chosen above has a radius of gyration of 6.28′′.This is more than double the assumption in the previous example. Instead of the

trial-and-error method used above, let’s determine the capacity of this pipe and adjust

from there.

Determine the capacity on an 18′′ × 1

4′′ wall pipe.

SR = [1.0 × 26′ × 12′′∕ft]∕6.28 = 50

Using Appendix 4, fb = 18.35 ksi


FIGURE 7.20 Aerial view of a strutted cofferdam.

If f = P∕A, then Pall = 18.35 ksi × 13.94 in2

Pall = 256 k

If the design called for a minimum capacity strut of 303 k as previously used,

this A36, 18′′ × 1

4′′ wall pipe would not be acceptable. The radius of gyration of this

pipe is substantial, but the cross-sectional area is small. Compare the normal stress

between the actual strut load and the actual capacity of the pipe:

fc =303 k

13.94 in2= 21.74 ksi

fc =256 k

13.94 in2= 18.36 ksi

This normal stress with either case is at or below the bending limit of 21.6–24 ksi

and below the compression limit of 28.8–32.4 ksi.

Determining the capacity of an axial loaded column or strut is another way to use

material that is available to the contractor. However, with this method, the available

strut will determine the spacing and the waler size. Therefore, this is not the preferred

method because most of the money is in the sheet pile and waler design. This method

is desirable when a designer just wants to know what a particular column or strut

capacity is. To determine the maximum axial capacity of a standard 12′′ diameter,

26-ft-long pipe, use Appendix 2 and follow these steps:

1. Find r = 4.38′′.

2. Find A = 14.6 in2.

3. Select the strut member 12′′ standard from Appendix 2.


4. SR = [1.0 × 26′ × 12′′∕ft]∕4.38 = 71.

5. Go to Appendix 4, fb = 16.33 ksi.

6. If f = P∕A, then Pall = 16.33 ksi × 14.6 in2.

7. Pall = 238 k

The conclusion is that a 12′′ standard pipe with le = 26 ft can support an axial

load of 238 k. If this pipe was supported with a brace, say in the center of its length

restricting movement 360∘, the capacity would change. The designer would have to

repeat steps 1 through 4 above and change the 26 ft in step 4 to 13 ft.

Web Yielding In some of this chapter’s photos, one might notice plates welded

against the beam web between the top and bottom flanges. These are stiffener plates

that are added when the strut (or any support member) force concentrates enough

load on the beam or waler that the web of that same beam yields (collapses) as shown

in Figure 7.21.

Allowable web yielding atress is a percentage of the steel’s yield strength, usu-

ally 75–80%. Using 75%, the allowable yield stress for A36 and grade 50 steel is as

follows:

Fwy = 0.75Fy

A36∶ Fy = 0.75(36 ksi) = 27 ksi

A572, Gr. 50∶ Fy = 0.75 (50 ksi) = 33 ksi

Web yielding stress is calculated by projecting the concentrated load into the top

flange of the supporting beam at a 2.5 : 1 angle each direction. Figure 7.22 shows this

FIGURE 7.21 Web yielding diagram.


FIGURE 7.22 Web yielding zone.

2.5 : 1 projection from the bottom flange of the top beam downward in both directionsto the bottom beam. These two projected lines cross the imaginary horizontal line kdistance from the top of the bottom beam. The k value represents the distance fromthe top of the beam to where the web thickness begins to change (at the tangent)and can be found in the AISC steel manual under beam properties or Appendix 1.Because the projected 2.5 : 1 line slopes each direction, the horizontal distance is5 times k.

The zone of concern is along the k horizontal line times the web thickness, whichis the area in square inches. In order to find the length of this area, the bottom flange ofthe top beam should be known. This is also found in Appendix 1 as the bf dimension.The web thickness of the lower beam is tw.

Therefore,

fwy =P

(tw)[bf + (5k)]

where P = concentrated load from the beam above

tw = thickness of the lower beam’s web

bf = flange width of the upper beam

k = the lower beam

Assuming a load (P) of 133 k from a top W21 × 147 onto a lower HP 14 × 89,determine the web yielding stress of the HP 14 × 89:

Fwc =133 k

(0.615′′)[12.51 + (2 × 1.31′′)]= 14.3 ksi

14.3 ksi < 0.75Fy

14.3 ksi < 27 ksi (OK)


TABLE 7.1 Shoring Costs for Different Systems

System Type Low $/ft2 High $/ft2

Shotcrete and soil nailing $15 $20

Soldier beam and plate $15 $25

Soldier beam and wood lagging $18 $28

Sheet pile, walers & struts $25 $50

CDSM (Cement Deep Soil Mixing) $20 $50

System Costs When costs are compared between different supports of excava-

tion systems, the unit of comparison is cost per “exposed” square foot (SF). The

exposed area (SF) of a shoring system is the perimeter in feet times the depth from

the original ground to the subgrade or dredge line. In other words, the amount of

sheet pile or soldier beam below the bottom of excavation is not counted toward the

square footage. For example, if a cofferdam is 22 ft wide by 46 ft long and 19 ft deep,

the square foot of exposed surface is perimeter length times the excavation depth or

[(22 × 2)′ + (46 × 2)′] × 19′ = 2584 ft2. Table 7.1 gives the reader a rough idea of

shoring costs for different systems. Actual costs should be calculated using actual

project parameters. Even though some costs are much higher, these systems may be

chosen because the less expensive system is not practical. This could come from the

depth, water tightness requirements, or existing conditions.

CASE STUDY: SHEET PILE COFFERDAM

In San Jose, California, theWWTPwas being expanded in 2006. Because of the

tight job site conditions, much of the excavation required shoring so that real

estate was not lost from open-cut excavations. The project even used a tower

crane on rails to eliminate having to have enough room for crawler and rough

terrain type cranes.

One particular excavation was over 38 ft deep. The excavation required a

long-reach excavator, which is very slow when it comes to production. Also,

because even the long-reach excavator could not reach the whole excavation, a

smaller excavator had to be hoisted into the excavation to complete the subgrade

and help place the aggregate base rock at the bottom. Figure 7.CS1 shows this

excavator being hoisted into the excavation.

The excavation was eventually a three-tier waler and pipe strut supported

excavation. Figure 7.CS2 shows this same excavator working at the bottom of

the excavation, feeding the long-reach excavator material to remove.


FIGURE 7.CS1 Sheet pile cofferdam (courtesy of Kiewit Infrastructure West).


(continued)


CASE STUDY: SHEET PILE COFFERDAM (Continued)

When the excavation was complete, the walers were also used to support the

dewatering pumps and discharge lines (Figure 7.CS3).


CHAPTER 8

PRESSURE AND FORCESON FORMWORK AND FALSEWORK

The committee that oversees formwork design is the American Concrete InstituteCommittee 347 (ACI 347) and is made up of 30–40 members. Its findings and deci-sions are recorded and published in Formwork for Concrete by M. K. Hurd. This textis in its ninth edition in order to keep up with the advancements in technology andmaterials. This same textbook has been used as the main reference for these writings.

For these next three chapters, the term formwork will refer to (1) vertical formssupporting the hydrostatic pressures of the unhardened concrete and (2) horizontalforms and components supporting the gravity dead loads from the concrete and liveloads from personnel, equipment, and tools.

8.1 PROPERTIES OF MATERIALS

8.1.1 Unit Weights

The unit weight of concrete will be 150 pcf for hydrostatic conditions. For falsework,the unit weight of concrete will be increased to 160 pcf to include additional weightof the reinforcing steel and formwork, not including steel beams.

The additional 10 pcf from reinforcing steel, form weight, comes from an averageamount of reinforcing used in structural concrete designs. Here are some sample cal-culations when determining this additional weight, which begins with an estimate ofhow many pounds of rebar are in one cubic yard of concrete. The additional 10 pcfincludes reinforcing. Table 8.1 shows sample weights to provide a guide in case thereinforcing steel is excessive and the engineer needs to add more weight for unusualcircumstances.

From Table 8.1, one can adjust the unit weight of concrete on falsework due to theamount of steel per cubic yard if 10 pcf is exceeded. This should be checked with a

155

156 PRESSURE AND FORCES ON FORMWORK AND FALSEWORK

TABLE 8.1 Rebar Weight per Cubic Yard [CY(yd3)] ofConcrete

Amount of Rebar

Pounds of Rebar

(lb/CY)

Pounds per Cubic

Foot (pcf)

Mildly reinforced 150 5.6

Light 175 6.7

Medium 200 7.4

Heavy 300 11.1

Very heavy >500 >18.5

registered engineer, but in no case would the 160 pcf be reduced. A simple quantity

takeoff can be performed to determine how much reinforcing steel is actually in the

elevated slab or supported concrete element. A complete takeoff is not necessary.

Simply select a typical section (20′ × 20′) and calculate the weight of the steel in that20 × 20 section, and then divide that number by the volume of concrete in the same

section.

Example 8.1 Determine the amount of steel in a 20′ × 20′ section of elevated slab

with a thickness of 12 in and No. 6 bars top and bottom at 12′′ on center, each direc-tion. The No. 6 bars weigh 1.502 lb/lf per Table 8.2.

TABLE 8.2 Reinforcing Steel Weight

Bar Size

Designation (No.)

Nominal Bar

Dimension (lb/ft)

Nominal Weight

(lb/ft)

3 0.375 0.376

4 0.50 0.668

5 0.625 1.043

6 0.75 1.502

7 0.875 2.044

8 1.00 2.670

9 1.128 3.400

10 1.270 4.303

11 1.410 5.313

14 1.693 7.650

18 2.257 13.60

Source: Erico International Corporation (Lentun®), 34600 Solun Road,

Solun OH 44139.

Reinforcing Steel:

Top 20 ft long × 21 bars × 2 directions × 1.502 plf = 1262 lb

Bottom 20 ft long × 21 bars × 2 directions × 1.502 plf = 1262 lb

Total = 2524 lb

8.1 PROPERTIES OF MATERIALS 157

Volume of Concrete:

20′ × 20′ × (12′′∕12′′∕ft)∕27 ft3∕yd3 = 14.81 yd3

Amount of steel = 2524 lb∕14.81 yd3 = 170.4 lb∕yd3

Based on the figures above, the unit weight for this slab would be 150 + 6 =156 pcf. Since the value is less than 160 pcf, the value used for weight would remain

160 pcf.

8.1.2 Forces from Concrete Placement

In this chapter, pressure formulas will be used and referred to as formula 1 andformula 2. These two formulas will be described in detail at that time.

Hydrostatic Pressure on Formwork The pressure that is imposed on verti-cal forms can never be overestimated. Since concrete weighs approximately 150 pcf(varies with mix design), the forces that are generated while it is in its liquid state canbe extremely high. Concrete remains in a liquid state for a short period of time, butduring this period, due to its unit weight, it is very heavy and can produce very highlateral pressures on formwork. Concrete is also unique to soil and other constructionloads as it is liquid for a relatively short period of time before it begins to take an ini-tial set and eventually (after a few hours or days) is able to support itself. The initialset of concrete occurs when the concrete is no longer imposing lateral pressure onforms. In terms of falsework loads, the initial set time does not lessen the forces ofgravity on a falsework system. In this case, the concrete must support its own deadload before forms can be removed. Other solid materials, such as wood and steel,maintain the same forces throughout its useful life. Even soil, other than when thewater table affects its stability and unit weight, is a relatively consistent force.

When the hydrostatic pressure of concrete is in question, several factors have tobe considered. Among these is the ambient air temperature and concrete temperatureat the time of the pour, the rate at which the concrete is placed, the mix design of theconcrete including its slump, the way in which it is placed and consolidated, and theheight of the placement. Let’s look at each one of these factors individually.

Concrete Temperature and Ambient Air Temperature For formwork design,temperature has a large effect on the lateral pressure on formwork. The warmer thetemperature, the faster the concrete will take its initial set. To the contrary, the colderthe temperature, the longer it takes the concrete to set. It is common for form fail-ures to take place in the winter months. This is because the form designer may havedesigned the forms in the warmer months and underestimated the temperature at thetime of anticipated construction. If the form designer designed the forms in Augustand the ambient temperature at that time was 90∘F, he may have used 60–80∘F asthe controlling temperature in formula 1 or 2. However, if the work is performedin January and the ambient temperature is recorded below 30–40∘ this could bringthe concrete temperature down below 50∘F, which would put more pressure on theformwork than was originally planned.


When temperatures are expected to be either very high or very low, batch plantstake measures to reverse the effects of either one. For instance, ice or liquid nitro-gen can be added to the mix to cool down the concrete temperature. If the concreteis being placed in cold weather conditions, the batch plant can add hot water tothe mix to attempt to raise the temperature of the concrete. In either case, the formdesigner should consider these attempts at temperature manipulation and how theywill affect the lateral pressure of the concrete. These effects are also increased whenadverse temperatures are combined with the use of flyash or retardant admixtures.Flyash is a by-product of the coal firing process. The finely divided residue is usedin ready-mixed concrete and can replace a certain percentage of the specified cementcontent by weight. Many mix designs will accept up to 40% of the cement contentas flyash. Admixtures are chemicals added to a mix for various reasons. Two of thereasons are better workability and slower set time.

When ACI 347 discusses temperatures for formula 1 or 2, it references the “con-crete temperature in Fahrenheit degrees in the forms.” This is tomean that the ambientair temperature could be 60∘F, but the concrete in the forms is over 90∘F. This posestwo problems: (1) the temperature in the formula would lower the maximum pressurepossibly more than ACI 347’s intentions, (and 2) most concrete mix design specifica-tions do not allow concrete to get warmer than 80–90∘F. If the concrete temperatureis going to reach temperatures above the specifications, the concrete supplier musttake measures to lower the temperature. In this case, the form designer will considerthis and make appropriate adjustments.

CASE STUDY: THE BENICIA–MARTINEZ BRIDGE—CONCRETE

TEMPERATURE

The Benicia–Martinez Bridge project (2000–2007) required mix designs with

up to 11 sacks of cement per cubic yard of concrete and a design strength of

11,000 psi. This amount of cement was unprecedented. The concrete temper-

ature was exceeding the specification limitations and peaked at approximately

190∘F (water boils at 212∘F at normal atmospheric pressures). Excessive tem-

peratures during the concrete curing process are known to shorten the long-term

strength and durability of the concrete. Excessive variances in temperatures

within a singlemass could also cause significant thermal cracking and reduction

in quality. The project required that cooling measures be installed to prevent the

concrete from reaching a maximum temperature threshold, as well as a maxi-

mum temperature variance. Measures included injecting liquid nitrogen at the

batch plant to precool, and cooling tubes were installed within the structure

to continue even cooling during curing. Cooled water was pumped through

the cooling tubes to keep the temperature as low and consistent as possible.

Even with those measures in place, the concrete temperature was still recorded

to peak above 140∘F. The following Figures 8.CS1 and 8.CS2 illustrate this

project under construction and completed.

The State of California and the contractor eventually had to accept this

“best attempt” at lowering the temperature and maintaining the temperature


FIGURE 8.CS1 Benicia–Martinez Bridge under construction looking from Benicia,

California (courtesy of Kiewit Infrastructure West).

FIGURE 8.CS2 Benicia–Martinez Bridge completed looking from Martinez, Cali-

fornia (courtesy of Kiewit Infrastructure West).

after placement so the differential between the outside temperature and the

concrete was not excessive. To monitor how well these attempts worked,

the contractor employed real-time measurement of the concrete temperature

from the moment of batching, throughout the initial concrete curing phase,

(continued)


CASE STUDY: THE BENICIA–MARTINEZ BRIDGE—CONCRETE

TEMPERATURE (Continued)

when temperatures reach their peak. Once concrete was placed in the truck,

handheld thermal-sensing devices were used to check the temperature during

transport. Within the concrete structure, thermocouple wiring was installed

within the rebar framework and connected to recording devices. These devices

monitored and recorded the temperatures at multiple locations within the

poured structure. Workers adjusted the water flow and temperature as neces-

sary to maintain proper cooling on a 24/7 basis. If any recorded temperatures

or variances exceeded the maximum recommendation, the contractor and the

state performed an engineering analysis to decide whether that portion of

concrete could remain in place.

Concrete cooling work at the Benicia Bridge required over 1.2 million feet

of PVC pipe and over 700,000 ft of thermocouple wiring.

Now if the form designer on the Benicia–Martinez Bridge had used 140∘ inits maximum pressure calculations (either formula 1 or 2), Pmax could be lessthan 600 psf, the forms would be well underdesigned, and there would be a highpotential for disaster. The design pressure should be discussed and agreed upon witha registered engineer on what temperature to use in the pressure calculations.

Placement Rate Placement rate is how fast the concrete is placed and vibrated inthe forms. The value is measured in vertical feet per hour of placement and designatedby R. Placement rate is typically decided on and controlled by the contractor. Thisdecision is communicated to the form designer for accurate pressure calculations.It would not make sense for the designer to design the forms for a pour rate of 4 ftper hour and the contractor to place the concrete at 8 ft per hour because the formswould be overloaded and bending, shear, and/or deflection on the forms would bejeopardized. On the other hand, it would not make sense for the form designer todesign at a rate of 6 ft per hour when the contractor can only place, at best, 4 ft perhour. In this case, the forms would be overbuilt and not as economical ($/SF).

Placement rate is controlled by many factors. A few of those factors will be dis-cussed here. Access to the forms or to the pump is crucial. The faster the mixer truckscan get to the pump, the faster the pump can provide concrete to the forms. If the con-tractor can continuously keep two trucks at the pump for the duration of the pour, moreconcrete can be supplied to the forms at a faster rate. To do this, there has to be plentyof space for the two trucks and room for turning around. Sometimes small earthenramps at the pump’s hopper can raise the mixer trucks discharge just enough to speedup the flow of the dispersed concrete. Crane and bucketing concrete, when pumpsare not economically or geometrically feasible, is an option and controls the rate atwhich concrete is placed. Buckets range in size from 1 to 4 yd3 and larger. A cubicyard of concrete weighs approximately 4050 lb. Therefore, crane capacities for thismethod should be substantial since a 4-yd3 bucket full of concrete could weigh over17,000 lb (8.5 tons).


The thickness of the elements being placed will have a very large effect on how

fast the placement rate is. For instance, an 8-in-thick wall will pour much quicker

in vertical feet per hour than a 24-in-thick wall. Let’s compare the two walls if they

were both 12 ft tall and 40 ft long:

Wall 1∶ 8′′ × 12′ × 40′

Volume = 11.9 CY

The amount of vertical feet placed by a 9.5-CY load of concrete 9.5∕[(0.67′ ×40′)∕27 CF∕CY] = 9.5 ft

This would result in less than half a truckload to place at 4-ft per hour rate.

Wall 2∶ 24′′ × 12′ × 40′

Volume = 35.7 CY

The amount of vertical feet placed by a 9.5-CY load of concrete 9.5∕[(2.0′ ×40′)∕27 CF∕CY] = 3.2 ft

This would result in a little over one truckload to place at a 4-ft per hour rate.

As you can see from this example, the thickness of the element is crucial to the

estimated placement rate. This puts a huge responsibility on the placement crew and

the engineer ordering the concrete. The concrete delivery speed must meet the place-

ment rate so there are no mistakes in this rate, and forms are not overloaded per the

design.

Mix Design Specifications will go into great lengths regarding the components and

their quantities in a cubic yard of concrete. Cement types, aggregate sizes, and admix-

tures are among a few specifications that can change the unit weight of concrete and

the lateral pressure on forms. ACI 347makes reference to these items and their effects

on form design.

The design formulas used in this chapter assume type I cement is used, the slump is

less than or equal to 4 in, internal vibrators are used, and no pozzolan or other retard-

ing admixtures are used. Pozzolan can possess cementitious qualities when mixed

with water and calcium hydroxide. The higher slump, the more liquid the concrete

is, thus increasing the pressure and the time before initial set occurs. As mentioned

previously, set retarders keep the mix more workable for longer periods of time while

still achieving the same strength requirements intended for the design of the structure

by the engineer. Workability of the concrete is essential for the labor force placing it

and for proper consolidation of the concrete ingredients, particularly in cases where

reinforcing steel is very congested.

Placement and Consolidation Methods ACI 347 recognizes that the most com-

mon way to consolidate concrete in forms is with the use of internal vibrators. These

vibrators produce the frequency and amplitude necessary to consolidate the concrete

into a solid mass without increasing the lateral pressure above an acceptable amount.

Spading of concrete actually produces a slightly smaller pressure, while external

air or electric vibrators induce an extreme stress to the forms by shaking the form


components so much that they can be stressed beyond their initial intent, which isresisting the lateral pressure of the concrete.

Placement Height In recent editions of ACI 347, the committee updated their pres-sure formula to include a limit on the height of an individual wall. They determinedthat a form higher than 14 ft should utilize a more conservative pressure calculation.Standards now state that if the placement height is greater than 14 ft, the form designermust use formula 2. The pressure calculations later in this chapter take into accountthis change.

Placement heights (form heights) can also affect the way a form is constructed. Thespacing of the form components is determined by design calculations on individualbeam members. When a form height is over 18–20 ft, additional considerations needto be addressed to make sure the form can support itself vertically without unneces-sary deflections or undulations.

Adjustments to Lateral Pressure Adjustments to the lateral pressure from con-crete itself and how much pressure is applied to the forming system are commonlyused in formwork design. Two factors are most important: unit weight coefficientadjustment (Cw) and chemical coefficient adjustment (Cc). The unit weight adjust-ment is used when the unit weight of concrete is lower or higher than the 150 pcfassumed. Typically, this factor will lower the unit weight of concrete by 1–10 pcfinstead of raising the unit weight. The unit weight of concrete rarely, if ever, risesabove 150 pcf; however, when a lightweight aggregate is used, the unit weight canbe lowered.

The chemical adjustment is for different types of cements used in the mixdesign in conjunction with any types of retarders used. Different cement types willchange the way the concrete reacts, changes the cohesion between ingredients, andaffects the pressure on the forms. Retarders include any admixtures that delay thesetting of the concrete. This includes, but is not necessarily limited to, water reduc-ers, mid-range water reducers, high-range water reducers, and superplasticizers.Tables 8.3 and 8.4 show some example adjustments that would be used for both unitweight and chemical coefficients.

With the addition of these adjustments, the lateral pressure formula used will be

Pmax = (Cw)(Cc)(lateral pressure)

TABLE 8.3 Unit Weight Coefficient (Cw)

Unit Weight of Concrete,

w (pcf) Formula Where w = unit wt Adjustment (Cw)

Less than 140 0.5[1 + (w∕145 lb∕ft3)],not less than 0.80

0.80–0.99

140–150 None 1.0

More than 150 w∕145 lb∕ft3 >1.0

Source: American Concrete Institute.


TABLE 8.4 Chemical Coefficient (Cc)

Cement Type Adjustment (Cc)

Types I, II, and III without retarders 1.0

Types I, II, and III with a retarder

Other types containing less than 70% slag or 40% flyash without retarders

1.2

Other blends containing less than 70% slag or 40% flyash with a retarder

Blends containing more than 70% slag or 40% flyash

1.4

Source: American Concrete Institute.

CASE STUDY: THE FOLSOM DAM—MIX DESIGN

The Folsom Dam Phases I–IV involved lowering the main gates of the original

structure, adding a new auxiliary spillway chute, and improving other aspects

of the dam in order to allow the lake level to be lowered and ultimately take on

more water from its watershed area.

The contractors involved with all the phases of construction had the chal-

lenge of designing formwork that would meet the challenging schedule require-

ments while supporting concrete mix designs that requiredmix set retarders and

high flyash weights. The mix design was not that flexible by specification and

was designed for workability and consolidation and not for formwork design

economy.

By ACI requirements, concrete lateral pressures should be increased by up

to 1.40 times if there is 40% or more flyash and retarders used.

The wall heights exceed 14 ft and in some cases were up to 60 ft tall and

could be poured as such, as long as the wall thickness did not exceed 5 ft

(Figure 8.CS3).

FIGURE 8.CS3 Folsom Dam (courtesy of Kiewit Infrastructure West).

(continued)


CASE STUDY: THE FOLSOM DAM—MIX DESIGN (Continued)

The form designers on site, calculated a 6-ft pour rate at 60∘F concrete tem-

perature for a wall greater than 14 ft in height.

Before the chemical coefficient (Cc) was included, the lateral pressure was

P = 150 + (43,400∕65∘F) + [2800(6 ft)∕65∘F] = 1076 psf

Since the mix design had over 40% flyash and included retarders, the full

1.40 chemical coefficient (Cc) had to be used. After the Cc was included, thelateral pressure was increased to

Pmax = P(Cc) = 1076 × 1.40 = 1507 psf

Forty percent more load added 431 psf to the form design.

As a result, the contractor’s plywood thickness could not be less than 3

4′′ – 1′′

and the stud spacing, depending on the plywood selected, could not exceed

9–10 in on center. Most important, depending on the tie size selected, ties were

reduced to 4-ft spacing or tie diameter had to be increased to 11

4′′ in diameter.

The costs associated with the mix design choice were very high when form-

work was concerned. This does not mean that the overall decision was wrong.

It could be argued that the workability of the concrete and the increased resis-

tance to water velocity was well worth the cost incurred by the formwork.

Hydrostatic Pressure on Concrete Formwork Pressure on formwork acts

perpendicular to the face of the sheeting, which in most cases is plywood or thin steel

plate. The form designer must know the temperature, T, of the concrete at the time

of placement, the average rate, R, of pour and how tall (h) the wall or element will be

for this individual placement. ACI 347 has developed two formulas to represent two

distinct conditions.

Formula 1 For formwork less than 14 ft (h) in height and a pour rate (R) of less than7 ft per hour uses the following formula:

Pmax = CwCc[150 + 9000(R)∕T].Not to exceed 2000 psf.

Not to exceed P = 𝛾h (concrete unit weight × height of placement).

Minimum pressure to be used in any case is 600 psf.

Formula 2 For formwork greater than or equal to 14 ft (h) in height OR a pour rate

(R) of greater than 7 ft per hour uses the following formula:

Pmax = CwCc[(150) + (43,400∕T) + (2800R∕T)].Not to exceed 2000 psf.




Column Forms For columns less than or equal to 15 ft in height, use the following

formula:

Use formula 1:

Pmax = CwCc[150 + 9000 (R)∕T].Not to exceed 3000 psf.


where P = pressure on the formwork in psf

R = average placement rate in vertical feet per hour

T = average concrete temperature at time of placement

Cw = unit weight coefficient

Cc = chemical coefficient


The derivations of these formulas are not included as they are not published by

ACI 347. It should be noted that there is a minimum and a maximum lateral pressure

in all cases. For columns greater than 15 ft, use formula 1.

Pressure Diagrams Pressure calculations and diagrams are extremely useful in

determining the shape and dimensions of the pressures on the forms and at which

point in the form’s height the pressure begins to taper down to zero pressure, which

always occurs at the top of the placement (top of form). To draw a proper pressure

diagram, follow these steps:

1. Calculate Pmax, the lateral pressure using formula 1 or 2 (considering T ,R, andmix designs).

2. Draw the form height as a vertical line and pressure as a horizontal line (it is

preferred to draw to the same scale).

3. Add a diagonal line from the top of the form to the maximum pressure, which

indicates P = wh, full liquid pressure.

4. Draw a vertical line from the maximum lateral pressure (Pmax) derived in step 1to the diagonal line represented by P = wh. Make sure you use the lesser of the

three values described earlier in this chapter.

5. Calculate the distance from the top of the form to the point at which the

max pressure begins its decline to zero pressure. This distance is equal to

Pmax∕150.6. Label the form height, the P = wh, the max design pressure (P), and the dis-

tance from the top of the form to the point where the max pressure begins to

reduce.

Example 8.2 Draw a pressure diagram for a 15′ high form (h), concrete placed at arate of 5′ per hour (R), and at a temperature of 70∘F (T).


Since h > 14 ft, use formula 2

P = 150 + 43,400∕70∘F + 2800(5 ft)∕70∘F = 970 psf

(150 + 620 + 200 = 970 psf)

This value is less than 2000 psf and Pmax = 150(15′) = 2250 psf; therefore,

970 psf is the designed pressure.

Figure 8.1 shows how this pressure diagram would be drawn.

FIGURE 8.1 Example pressure diagram.

To calculate the dimension from the top of the placement to the maximum pressure

value (top of the vertical), divide the maximum pressure by 150 pcf.

y =Pmax

150

y = 970

150= 6.47 ft

Example 8.3 Draw a pressure diagram for a 12′ high form (h), concrete placed at arate of 6′ per hour (R) and at a temperature of 80∘F (T).

Since h < 14 ft and R < 7, use formula 1.

P = 150 + 9000 (6 ft)∕80∘ F; P = 825 psf

(150 + 675 = 825 psf)

This value is less than 2000 psf and P = 150(12) = 1800 psf; therefore, 825 psf is

the designed pressure.


Figure 8.2 shows how this pressure diagram would be drawn.

y =Pmax

150

y = 825

150= 5.5 ft

FIGURE 8.2 Example pressure diagram.

Full Liquid Pressure There are several conditions where lateral pressure should

be calculated using the full liquid pressure formula, P = wh, where P is the pressure

in psf, w is the unit weight of the concrete, and h is the form height. Some conditions

that require full liquid pressure are as follows:

Use of external vibrators (internal vibration) is considered normal.

Use of admixtures that slow the initial set of the concrete such as superplasticizers,

mix retarders, and newly introduced admixtures (when little field experience is

present).

Concrete that is pumped from the bottom using the pressure of the pumping device.

In some cases, 25% should be added due to the pressure from the pump piston.

This includes tunnel forms.

Slumps higher than 4 in.

When revibration of the freshly placed concrete is required.

Figure 8.3 showswhat a pressure diagramwould look likewith full liquid pressure.

The diagram remains triangular.


FIGURE 8.3 Full liquid pressure diagram.

Cantilevered Formwork When lifts of concrete are placed on top of previously

placed lifts of concrete, such as during dam construction when monoliths are placed

on monoliths, concrete loads and the type of forms used can introduce design

challenges. Figure 8.4 shows a monolith form ready for the next placement lift and

anchored to the previous lift of concrete with a Williams pigtail anchor. It is common

in these cases to attach the bottom of the form to an anchor placed in the top of the

previous pour. The anchor used to make this attachment has to be strong enough to

resist the force at the bottom of the pour, and the previously placed concrete has to

FIGURE 8.4 Cantilevered formwork anchored to previous lift.


FIGURE 8.5 Monolith dam construction cantilevered form.

have achieved adequate strength prior to the placement of the next lift. Usually this

would require the lower concrete to have at least 3000–5000 psi in strength.

The pressure diagram in a case like this could either be a full liquid head, triangular

diagram, or a trapezoidal diagram from formula 1 or 2. This would depend on the

same parameters mentioned earlier. If the form designer thinks the lift can be poured

very quickly (within one hour) or one or more of the full liquid head criteria apply,

then he would create a triangular diagram. If the form designer recognized that, based

on the amount of concrete needed for the lift, a particular placement rate will not be

exceeded; he can then develop a diagram based on formula 1 or 2.

The statics involved in this type of problem are relatively simple.

Sample Cantilevered Form Pressure A 6-ft concrete lift will be placed on a pre-

vious lift (like a dam monolith). A basic form will be used and a steel vertical waler

will support the form at 5′ 0′′ on center horizontally, as shown in Figure 8.5.

Example 8.4 Refer to Figure 8.6 and determine the force (T) in the anchor and theforce (C) in the compression point at the bottom of the waler. Assume full liquid

pressure from the 6-ft lift. This could be because the concrete could be placed in one

hour or because of one of the five reasons stated above when full liquid head should

be considered.

Determine the load diagram. In this case, it is assumed full liquid head. Therefore,

the pressure diagram shown in Figure 8.6 will be used. This is a triangular shape from

zero pressure at the top to a Pmax value at the bottom. Since the lift is 6 ft,

Pmax = (1.0)(1.0)(150 × 6) = 900 psf


FIGURE 8.6 Form drawing of cantilevered formwork.

Note that this value is for 1 ft of horizontal form only.

Calculate the resultant (R). For a triangular pressure diagram, the resultant is

located one third of the height from the bottom (1

3of 6 ft = 2 ft from the bottom).

The resultant of a triangular pressure diagram is the area of the triangle (b × h∕2).

R = 900 psf × 6 ft∕2 = 2700 plf of horizontal form

The resultant value can now be multiplied by the waler spacing of 5 ft. It can also

be done later in the problem, but it is recommended that it be done now.

2700 lb × 5 ft on center = 13,500 plf of wall

This is the resultant value on one waler spaced at 5 ft on center.

Now we have three reference points: Resultant (R), compression shoe (C), andtension anchor (T). Since the resultant has just been calculated, we have one known

value and two unknowns, which means we can total the moments about one of the

unknown points and solve for the second unknown point. Let’s add the moments

about (T). In order to do this, the distances need to be known.

(13,500 × 2.5 ft) − (C × 5 ft) = 0; therefore C = 6750 lb

Now add the forces in the horizontal direction.

13,500 + 6750 − (T) = 0; therefore, (T) = 20,250 lb

Draw a shear and moment diagram similar to Figure 8.7.

The FBD showswhat the required load on the embedded anchor needs to be. A fac-

tor of safety of at least 2.0 would need to be used. This would require an anchor with

a safe working load of 42 k. This is a rather heavy force for which to design. If there

is not an anchor that has this capacity, the designer can adjust the criteria. The first


FIGURE 8.7 Free-body, shear, and moment diagram of Example 8.4.

might be to close the spacing of the vertical walers in order to reduce the amount of

resultant load on one waler. The second option would be to add an anchor at the top

of the form. This would change the whole problem and the statics would then have to

be recalculated. This would definitely reduce the load on the bottom anchor while the

upper anchor would have to assume some of the load. The challenge would be to find

an anchor location for an upper tie. The last option might be to change the pour rate,

which would give us a different pressure diagram with a lower maximum pressure.

This is not always a good option as it increases the labor cost; and if there are many

placements with this form, changing the pour rate could be very costly over the life

of the project.

From the shear and moment diagram in Figure 8.7, the designer can determine the

size of the vertical waler. From the maximum moment of 33,750 ft-lb, the minimum

required section modulus can be determined, and a beam can be selected. The fol-

lowing two examples illustrate double waler selection. A double waler is used so that

the anchors can pass through the gap between the beams and concentrically load the

two beams as illustrated above in Figure 8.5.

Example 8.5 Double steel waler using American channels. Determine the beam

size based on bending only (shear will not be checked due to use of steel).

Determine the type of steel and the factor of safety (FOS). Assume A36 steel with

a 1.5 : 1.0 FOS.

Fb =Fy1.5

therefore

Fb =36,000 psi

1.5= 24,000 psi

Calculate the minimum section modulus required for a single channel.

S = MFb

therefore,

S =33,750 ft-lb × 12′′∕ft

24,000 psi= 16.875 in3


Determine the size of a single steel American channel. Theminimum sectionmod-

ulus needs to be divided by two since it is a double channel.S∕2 = single channel minimum section modulus, therefore, S = 16.875 in3∕2 =

8.44 in3.

Look up this channel in Appendix 1 with a section modulus equal to or greater

than 8.44 in3.

Conclusion: Use an American channel size C 9 × 13.4.

Example 8.6 Double, wood 4 × waler using Douglas fir with Fb = 1800 psi and

Fv = 150 psi. Since we are using wood walers, shear will also need to be checked.

First select the material size based on bending requirements and then check the shear

strength of that same size beam. The beam selection has been reduced to 4 × material

only from experience.

The material has been determined and the values already include FOS.

Calculate the minimum section modulus of the double waler beam.

S = MFb

therefore,

S =33,750 ft-lb × 12′′∕ft

1800 psi= 225 in3

Determine the size of a single Douglas fir waler. The minimum section modulus

needs to be divided by two since it is a double channel waler.S∕2 = single channel minimum section modulus, therefore, S = 225 in3∕2 =

112.5 in3.

Assuming a base dimension of 3.5 (4 × S4S dimension) inches on one of the

beams, determine the minimum depth of the Douglas fir single beam.

S =bd26

therefore,

112.5 in3 = (3.5′′)(d)2∕6

Solve for d, d = 13.88 in (shear has not been checked in this example).

Assuming shear is not the controlling factor, next decide if this size beam is fea-

sible. Based on these calculations, this waler would have to be a Douglas fir 4 × 16.

Because this beam size is very large, the designer would most likely decide to change

some of the controlling parameters in order to use a smaller beam. This could include

reducing the vertical spacing or increasing the minimum dimension to a 6×. Anotheroption would be to go with steel double channels as used in Example 8.5.

One-Sided Forms One-sided forms are used when a concrete wall is requiredagainst a shoring system, sloped earth, or vertical rock faces. In other words,one-sided forms are used when the concrete only has to be formed on one sidebecause the other side is up against a stable surface. Figure 8.8 shows a one-sided


FIGURE 8.8 One-sided form.

form attached to a sheet pile wall. One-sided forms are typically designed the sameas a two-sided form except the designer usually attempts to minimize the numberof ties. Ties for a system like this are typically more expensive than standard tiesbecause of the added cost in labor and materials to attach the tie to the shoring wallor rock face. As in Figure 8.8, each tie must be attached to a welded bracket againstthe sheet pile. The pressure on a one-sided form is calculated the same way as atwo-sided form. It is based on form height, pour rate, and temperature.

The formwork placement and removal costs for a one-sided wall are higher than atwo-sided wall form by approximately 40–60%,mostly due to the ties/anchors attach-ing the form to the other side. In order to reduce the cost of the one-sided form, thedesigner typically increases the size of the anchors/ties in order to reduce their quan-tity. When the number of rows are reduced, not only does the ties size increase butthe double waler size would also increase due to the spacing between ties becominglarger and increasing the maximum moment on that vertical or horizontal waler. Thiswill be discussed more in Chapter 9.

Concrete Pressure Underwater Concrete placed under water has different effectson formwork because concrete is lighter under water than on land. For the most part,the difference in unit weight between concrete under and above water is the differ-ence between the unit weight of concrete and the unit weight of the water (150 −62.4 pcf = 87.6 pcf). The pressure of the water at 62.4 pcf counters the concretepressure.

The placement heights are always different, but underwater placement can be sim-plified if the engineer assumes full liquid pressure at 87.6 pcf times the height ofplacement. Water depth (atmosphere) can affect this value, but it is negligible.

Example 8.7 Calculate the placement pressure under water on a form with concrete

6 ft high. Then draw a pressure diagram (Figure 8.9).

P = 6 ft × 87.6 pcf = 525.6 psf


P = 6 ft × 87.6 psf

6ʹ

FIGURE 8.9 Underwater pressure diagram.

CASE STUDY: CHRYSTAL SPRINGS PROJECT

A project in northern California called Chrystal Springs required concrete

placement under water up to depths over 100 ft. This project was upgrading

the infrastructure of the water supply system to the San Francisco Bay Area.

Included in this work was the demolition and reconstruction of several under-

water structures that were built in the late 1800s and early 1900s. Most of the

underwater structures that existed were built by brick and mortar construction.

(Figure 8.CS4).

FIGURE 8.CS4 Barge arrangement for underwater work. (courtesy of Kiewit Infras-

tructure West)


The project staff had to design forms to support the hydrostatic forces caused

by the underwater placement.

The formwork for underwater work was selected to be steel forms so that

buoyancy of lighter materials was not an issue. The forms selected were rented

from EFCO Corporation and came with all the bolts, plates, and other acces-

sories for attaching to one another.

Falsework Loads Elevated concrete loads are a combination of the dead load

(DL) of the concrete, the live Load (LL) from the tools and workers, and additional

weight of the falsework beams:

max = DL + LL + steel beams

where DL = slab thickness (feet) × 160 pcf (160 = concrete unit weight including

rebar and formwork weight, not including steel)

LL = depends on the type of construction (ranges from 25 to 75 psf)

Light duty = 25 psf (painters)

Med duty = 50 psf (carpentry)

Heavy duty = 75 psf (masonry)

Special = designed by a registered engineer

In this chapter, a flat slab concrete deck (Figure 8.10) and a T section of deck

(Figure 8.11) will be used in the examples to demonstrate basic falsework loads.

In Chapter 10, bridge falsework will be introduced as a more sophisticated problem.

FIGURE 8.10 Matrix diagram of falsework loading.

FIGURE 8.11 Elevated concrete in T sections.


Example 8.8 Calculate the design load of a 30′′ slab with medium live load.

DL = 30′′∕12′′∕ft × 160 pcf = 400 psf

LL = 50 psf

Total = 450 psf

Example 8.9 Determine the weight of a 1-ft-long T section shown in Figure 8.11

if the webs are equally spaced at 6 ft on center. The other properties of the structure

are as follows.

Overall height = 5 ft

Top slab thickness = 12′′ thick

Web = 16′′ thick

Top flange (6 ft × 12′′∕12′′∕ft)(1 ft)(160 pcf) = 960 plf of T

Web (5 ft − 1 ft)(16′′∕12′′∕ft)(1 ft)(160 pcf) = 853.3 plf of T

Total weight plf of T = 1,813.3 lb

When the falsework is designed for this elevated section, most likely one stringer

beam will be designated per T section. This stringer will have to support 1813.3 plf

plus the live load and its own weight, depending on the size of the beam. This will be

covered in detail in Chapter 10.

Formwork and Falsework Removal Project specifications typically control the

time frame between concrete placement and form removal. In addition, there are at

least two distinct conditions for form removal criteria. The first is how long a verti-

cal form must remain in place before its removal. This is determined by the amount

of time the concrete needs before it can support itself without the aid of side forms,

which controlled the initial lateral pressure. The second case covered in the specifi-

cations is the amount of time before falsework can be removed. As discussed before,

falsework supports the dead load of the concrete, the live loads from personnel and

equipment/tools, and falsework materials. The concrete is usually reinforced either

with rebar or a combination of rebar and posttensioning rods or strands. Therefore,

the time that elapses before the concrete, reinforcing, and/or posttensioning can sup-

port the elevated slab or bridge can be between 14 and 28 days. It is more common

for the engineer to require 28 days or when the concrete reaches its design strength,

which is typically between 4000 and 6000 psi. The engineer may also require that

both of these be met.

Posttensioning creates some situations that can increase removal time and apply

additional loads on the falsework that possibly were not planned for during the design

phase. In bridge construction, cure time of approximately 10–15 days must elapse

after the last placement before the posttensioning strand can be stressed to the design

loads. After tensioning takes place, then the falsework can be removed. The postten-

sioning typically follows a curved pattern, similar to Figure 8.12, where the strands


FIGURE 8.12 Post-tensioning pattern of a typical two-span bridge.

are high at the supports and low in the mid-spans. This allows the strand to lift up

the spans of the bridge, which in turn applies a downward force at the piers and abut-

ments. This increased downward force adds load to the falsework support members

located on each side of the piers and adjacent to the abutments, which could make

removal more difficult. At this time the bridge is self-supporting, so the effects on the

design of the falsework can be ignored.

The next two chapters will implement the concepts learned in this chapter as they

apply to formwork and falsework, respectively. The student will have several oppor-

tunities to apply loads and pressures to design situations used in the industry today.

CHAPTER 9

CONCRETE FORMWORK DESIGN

In Chapter 8, we covered the pressures and loads that are generated by liquidconcrete. From that information, the reader should understand that these forces canbe extremely large; and if not planned and designed for adequately, they can bevery costly as well as dangerous. This chapter will use the load information fromChapter 8 and begin to design the most economical system for the forming conditionat hand. Since this chapter relates to formwork only, hydrostatic pressure will be themain focus. Loads will be given to enable students to spend more time on the designprocess and decision making. Like most temporary structure decisions, there aremultiple solutions to formwork design. The challenge is to find the most economical.

This chapter will include several wood structure and composite designexamples that go beyond basic formwork design. In addition, there is a detaileddiscussion on formwork ratios and how they relate to form costs.

9.1 GENERAL REQUIREMENTS

9.1.1 Concrete Specifications

Specifications for concrete work are typically separated between formwork and cast-in-place concrete mix designs. The mix design of the concrete can occupymany spec-ification subsections, depending on the engineer. Formwork specifications give thecontractor guidelines while not attempting to require specific means and methods.Owners and engineers are usually reluctant to force a contractor into a particular wayto build unless the risk to the design is so great that they must intervene through theuse of the specifications.

Formwork specifications tend to be slightly different from engineer to engineer;however, specifications covering the most critical aspects of concrete formwork tend

178

9.1 GENERAL REQUIREMENTS 179

to be consistent. The following sections will illustrate some of the similarities fromone concrete specification to another. These items, at a minimum, should give thestudent a basic idea of what to look for in concrete specifications.

Submittals Submittals have always been a requirement for falsework design.CalTrans and other agencies may define submittal requirements based on these ele-ments: height, maximum span length, whether spanning water or existing roadways,and the like. Formwork design, on the other hand, is not always a required submittal.However, of late, more engineers are seeing value in formwork design reviews.Construction litigation and risk has driven the owners and engineers to take a moreproactive approach to contract management.

Materials Engineers may only allow particular form sheeting or not allow another.By doing this, they may be trying to assure that their standard of concrete quality ismaintained. Formwork ties are usually an item that is seen after completion, and thetype and spacing used could be a concern to the design engineer.

Loading It is fairly common that an engineer would suggest a minimum load onfalsework or formwork. The engineer may insist on a minimum live load (say 50 psf)and possibly a minimum dead load (say 100 psf). Anything below these two valueswould be designed at a minimum of 150 psf, combined DL + LL.

Execution of the Work There is usually a quality discussion so that the contrac-tor understands the expectations of the owner/engineer. This can easily be done byreferring to ACI 347 standards for quality. The manner in which the forms are built,placed, and removed may be specified. The length of time a form must remain inplace after concrete placement is typically specified as a minimum duration in daysor hours. The engineer would also reiterate the expectation of form cleanliness priorto concrete placement so that debris is not incorporated into the final product.

The location and distance between construction joints is critical to the water tight-ness of water-retaining structures, and these parameters would be limited by theowner/engineer. This could simply be to reduce the chances of cracking.

The number of times a form can be used may be specified, or at least a discussionwould take place concerning when a form is no longer acceptable for reuse.

California state or federal OSHA is usually referenced to the contractor so itis clear that safety standards will be maintained in accordance with the properauthorities.

Specifications will vary, but the previously mentioned items are some commonitems that are found in most general concrete provisions.

9.1.2 Types and Costs of Forms in Construction

Table 9.1 lists formwork types used in construction along with a range of the cost($/SF) for the formwork material, not including labor, equipment, and so forth. Notall these elements would be design considerations in this chapter, but all are integralto concrete construction in one form or another. Typically, an element becomes aconcern to form designers when it is elevated, very tall or thick in dimension, orunusual.

180 CONCRETE FORMWORK DESIGN

TABLE 9.1 Cost of Form Elements

Element Type of Form Estimated Material Cost per SF

Footing Wood, braced to soil $1.00–$1.50

Slab on grade Wood, braced to soil $0.75–$1.25

Pier foundation Steel or wood forms $2.00–$4.00

Small walls Hand set forms $2.50–$8.00

Large walls Gang forms $8.00–$30.00

Columns/piers/pilasters Steel, wood, and fiberglass $10.00–$15.00

Elevated slabs Shoring frames, wood post $2.00–$5.00

Bridge superstructure Falsework and soffit $3.00–$6.00

Bridge deck Carrier beam (horse) $2.00–$4.00

Construction joints Wood, throw away, stay-form $1.50–$3.00

Edge of deck Wood, braced to deck $1.00–$1.50

9.2 FORMWORK DESIGN

9.2.1 Bending, Shear, and Deflection

We have learned previously that beam design has to pass three different tests: bend-

ing, shear, and/or deflection. Unlike earth support, formwork has more restrictions

because the finished product is permanent in nature, thus exposed to the public eye.

In earth support systems, shear is not an issue for the steel components and the lag-ging are used flat, so shear is not the weak link. As for deflection, earth shoring can

deflect slightly without being a problem unless it causes excessive settlement. This

leaves bending as the most critical element. However, in formwork design, all threeof these tests are critical to the success of the form design. Shear becomes impor-

tant because wood is a common forming material. Deflection becomes important for

aesthetics.

Since formwork is simply a series of beams, each one has to comply with all threechecks to assure design adequacy. The three checks will be discussed in more detail.

Formwork members are also typically continuous over multiple supported members.

For instance, plywood spans multiple studs; studs span multiple walers, and so forth.Therefore, the formulas used are those that consider a continuous beam over more

than two or three supports. We call these indeterminate structures because there are

more than two unknowns. In some formwork design problems, when the loading

conditions are very complicated, beam analysis software is used to speed up the engi-neering process. When special loading conditions are encountered, standard static

practices should be used.

It should be noted that what we learn here with bending, shear, and deflection will

also be used in Chapter 10, where applicable.In formwork design, tables are used for material property information. The two

types of tables most commonly used for formwork are (1) specie information with

allowable stress values and (2) section property information, which is based on thebeam’s dimensions in cross section.

Specie information would give the engineer the allowable stress values for bend-

ing, shear, compression perpendicular and parallel to grain, and modulus of elasticity

9.2 FORMWORK DESIGN 181

at the least. These values may or may not be adjusted for the various adjustments

mentioned in Chapter 2 such as duration, wet use, flat use, and temperature.

Section property information would give the engineer information about the

dimensions of the sections, which would lead to the section modulus, the moment of

inertia, and the cross-sectional area. For plywood, the cross-sectional area is replaced

by rolling shear constant because plywood is a laminated product. The horizontal

shear strength is largely dependent on the glued laminates and a material testing

laboratory is generally used to determine these values from testing.

Bending Stress Bending stress is an important component of formwork. Since

bending stress is a derivative of bending moment and section modulus, our basic

formulas are used to solve for various unknowns. The following static and strengths

of material concepts derive the bending formulas used in formwork design. In most

cases we can solve for any variable. However, it is most common to solve for the span

of the supports knowing the material sizes and specie type.

Maximum bending moment (M), simply supported (two to three supports), one to

two spans, M = wL2∕8Multiple supports (more than two spans),M = wL2∕10Bending Stress, fb = Mmax∕S


L = span from the inside face of one support to the inside face of the

adjacent support

Section modulus (S)

Dimensional lumber, S = bd2∕6 from Appendix 6

Plywood (laminated), S = from Table 9.3


d = beam depth

Since most formwork is a series of beams over multiple supports, the second

bending moment formula shown above (wL2∕10) is more commonly used. In a

case where the designer knows there will only be two supports, the first formula

for bending would apply. With these three formulas, one can solve for any of

the different variables as long as the information is provided that leaves only one

unknown. For formwork, it is common to know what size material is desirable

and have to solve for the spacing of the elements. In this case, if we know all the

variables except L (span length), we can solve for the support spacing. Let’s see how

this would work.

Assume we are given the following:

Type of materials and their maximum bending stress values (Fb)

Uniform load on the beam in question (w)

Section modulus of selected materials based on size or thickness (S)


FIGURE 9.1 Span between studs.

We could then solve for L span by using the following formula. This formula isthe same for plywood and dimensional lumber:

L = 10.95√FbS∕w

Once the span (L) is determined, the designer knows the maximum spacing of thematerials supporting the beam section that was just analyzed. Figure 9.1 shows thespan dimension, which is the inside face of one stud to the inside face of the other.

The variables in the span formula for bending come from the statics and strengthformulas.

Shear Stress The same process can be used for shear stress, except the basic stat-ics and strengths of materials concepts of shear will be used. This is what we knowabout shear:

Maximum shear (V)

Simply supported (two supports), V = wl∕2 or 0.5wl

Multiple supports (more than 2), V = 0.6wl



adjacent support

Shear stress, fv = 1.5V∕ADimensional lumber, A = bd

Rolling shear for plywood (laminated), Ib∕Q


d = beam depth


Type of materials and its maximum shear values (Fv)

Uniform load on the beam in question (w)

Cross-sectional area or rolling shear of selected materials based on size or thick-ness (Ib∕Q) or b and d


Before we solve for L, ACI allows us to adjust the allowable shear stress by up to afactor of 2.0 for dimensional lumber. This basically means we can double the valuesgiven to us by the National Design Specification for Wood Construction. We callthe new shear value F′

v = 2.0 × Fv. For plywood, the value given by the plywoodassociation is used and does not get an allowance. This value is typically between 57and 75 psi.

Solve for L.Plywood formula:

L =Fv(lb∕Q)0.6w

Dimensional lumber formula:

L =(F′

vbd)0.9w

+(2d12

)Deflection Finally, the same process is used for deflection, with different variablescoming from the deflection parameters as follows:

Maximum allowable deflection for forms using L∕360Maximum deflection = 5wL4∕384EI (Simple span, two supports)

Maximum deflection = wL4∕1740EI (Multiple supports)



adjacent support

E = 1,300,000 psi to 1,700,000 psi depending on material

I = moment of inertia


Type of materials and its maximum modulus of elasticity (E) in psi

Uniform load on the beam in question (w) in PLF

Moment of Inertia (I) in in.4

Solve for L. The same formula is used for plywood and dimensional lumber.

L = 1.69 3√EI∕w

Load Path Load path for concrete formwork is not much different from other tem-porary structures. The path of the concrete pressure is as follows:

The concrete pressure is applied to the sheeting material (typically plywood, steelplate, or sometimes flat, dimensional lumber.

The sheeting loads the studs. Based on the stud spacing, the linear load on the studis determined.

The stud loads the waler system. Based on the spacing of the walers, the linearload on the waler is determined.

The waler loads the ties or struts. Based on their spacing, the total load of the tieor strut is determined.


Uniform Loads on FormMembers The most important aspect when followingthe load path is determining the uniform load on the next beam member to whichthe load is transferred. For instance, if it is determined that studs are to be spaced at14 in on center to support a plywood thickness, then the uniform load on the studmust take into account the 14-in spacing. The uniform load on the stud becomesPmax × 14′′∕12′′∕ft. If Pmax is 800 psf, then the uniform load on the stud is 933.3 plf.The unit is now pounds per linear foot since the spacing in feet is canceled.

Adjustment Factors

Adjustments to Material Stresses The wood industry has developed a numberof adjustment factors for permanent construction using lumber. ACI 347 has adopteda few of these, but not all of them, for temporary structures and formwork. Thesefactors are multipliers to either raise or lower the allowable stresses. For example, if aparticular factor is 1.20, then the allowable stress will be raised by 20%. Conversely,if the factor is more conservative, the multiplier would be less than 1.0, say 0.80;and in this case, the allowable stress would be reduced to 80% of its original value.More than one factor can apply to a single stress value. For instance, the short-termduration factor (for loads less than a few months) may increase the stress by 25%, butthe wet service factor (for material with moisture content more than 19%) may lowerallowable stress to 90%. Therefore, an allowable stress of 1200 psi × 1.25 × 0.90 =1350 psi.

Also, some adjustment factors only apply to some stresses. For example, oneadjustment factor may apply to bending stress and modulus of elasticity but not shearstress, whereas another factor may apply to shear stress only. Table 9.2 lists the mostcommon adjustment factors to permanent wood structures and temporary structures.

For the purpose of this textbook, three main adjustment factors will be used:

Load duration 1.25

Shear adjustment 2.0

Wet service factor 0.67–0.95

The stress values used in this chapter’s examples include the adjustment factorsalready.

TABLE 9.2 Adjustment Factors

Factor Symbol Applies to

Load duration Cd Fb, Fc, FvWet service Cm Fb, Fc, Fv,ETemperature Ct Fb, Fc, Fv,ERepetitive member Cr FbShear adjustment C′

v FvFlat use Cfu FbSize factor Cf Fb,FcBeam stability Cl FbColumn stability Cp Fc


Form Design Using Statics and Strengths of Materials The first type ofform design discussed in this text will be what is considered form design throughthe use of statics and material standard principles. This means we will use all thebasic statics and strengths concepts mentioned earlier in this chapter and apply it toan all-wood form design. Later in this chapter, steel and aluminum components willbe introduced for larger type form systems. Once the designer knows what type andwhat size of material is available or preferred by the contractor, the three formulasfor shear, bending, and deflection will be applied to determine the most effectivedesign.

In order to begin this process, select the material type for the followingcomponents:

Form Plywood—Standard BB-OES (both sides have type B finish, and the plywood has been oiled, edged, and sealed) Form Ply

Rolling shear strength (Ib∕Q)Bending stress (Fb)

Modulus of elasticity (E)

Dimensional Lumber—Stud and Waler Size

Shear stress (F′v)

Bending stress (Fb)

Modulus of elasticity (E)

Tie Material and Type

Tensile Strength

Innumerable species of wood are available throughout the world. However, fortemporary structures, the designer considers the use and selects a particular speciesof wood that is more affordable in the project’s geographical area. Common woodspecies for temporary structures are Douglas fir, Hem fir, and Southern Pine. Table 9.3lists estimated stress values for the more common lumber used for formwork. Thesevalues were derived from the NDS literature.

TABLE 9.3 Common Wood Specie Allowable Stresses for Formwork (psi)

Specie

Bending,

Fb

Compression

Perp-grain,

Fc∥

Compression

Parallel-grain,

Fc⟂

Horizontal

Shear,

Fv

Modulus of

Elasticity,

E

Douglas fir–larch 1000 625 1600 95 1,500,000

Douglas fir–south 1250 585 1600 90 1,600,000

Southern pine 1250 565 1600 90 1,200,000

Spruce–pine fir 975 405 1500 70 1,400,000

Hem-fir 975 405 1500 75 1,300,000

BB-OES plywood 1600 210 on face 60 1,500,000

Source: National Design Specification (NDS)


Table 9.3 includes important stress values that are used at one time or another

during form design. The following allowable stresses used for basic form design:

Shear stress, FvBending stress, FbModulus of elasticity, E

Some additional stresses used when analyzing members in compression are par-allel to grain (Fc∥) and perpendicular to grain (Fc⟂).

Plywood Design When deciding on plywood thickness, the designer can follow

one of three methods: (1) Select a particular thickness and type of plywood (oftenbased on experience), (2) select material owned by the contractor, and (3) select

material that will meet specifications. (For example, the concrete specifications mayalso lean the designer toward a smoother finish or a strict deflection requirement.)

Figure 9.2 shows how excessive deflection in plywood and studs can be out of spec-ification tolerance.

For the less experienced form designer, not all thicknesses of plywood are used.Plywood thicknesses range from 1∕4′′ to over 1′′ in thickness. For form design though,

most likely 5

8′′ to 3

4′′ will be the form ply thickness of choice. This text will use one of

these two sizes. Plywood also has alternating laminates, usually three, five, or seven

ply. The more plies, the better the quality of plywood. The number of plies is an oddnumber because each face of the plywood always has the grain of that ply running

in the direction on the length of the sheet. When using plywood as a form sheetingmaterial, it is best, whenever possible, to place the plywood in its strongest direction.

This occurs when the outside face grain of the plywood is running parallel to thespan of the plywood. Therefore, the weak direction of the plywood is when the face

grain is perpendicular to the span of the plywood. A common mistake is to interpretthese two options as parallel or perpendicular to the “stud direction.” This is a costly

mistake and should be avoided whenever possible. Most plywood charts give figures

FIGURE 9.2 Plywood and stud deflection.


FIGURE 9.3 Plywood spans, weak vs. strong direction.

for both cases and allows the designer to select the one that fits his design. Figure 9.3

illustrates this concept of plywood direction.

Most Commonly Used Plywood Types in Concrete Formwork

BB-OES

Both sides type B sanded face

Oiled, edged, and sealed

One side medium-density overlay (MDO)

One side high-density overlay (HDO)

span calculations on plywood Bending stress for plywood:

L = 10.95√FbS∕w

Shear stress for plywood:

L =Fv(lb∕Q)0.6w

Deflection for plywood:

L = 1.69 3√EI∕w

In order to use these span formulas, the form designermust havematerial allowable

stress values and property values as mentioned earlier. Tables 9.3 and 9.4 include esti-

mated values for plywood and dimensional lumber. These values have been estimated

from NDS and ACI.

Table 9.4 contains properties for moment of inertia, section modulus, and rolling

shear (Ib∕Q). Respectively, these values are used for deflection, bending, and shear

calculations. The table has a strong section and a weak section. As mentioned earlier,


TABLE 9.4 BB-OES Plywood Properties

Strong Direction Weak Direction

Thickness

(in)

Minimum

Ply I (in4) S (in3)Ib∕Q(in2) I (in4) S (in3)

Ib∕Q(in2)

Weight

(psf)

3

83 0.027 0.125 3.088 0.002 0.023 3.510 1.1

1

23 0.077 0.236 4.466 0.009 0.087 2.752 1.5

5

85 0.129 0.339 5.824 0.027 0.164 3.119 1.8

3

45 0.197 0.412 6.762 0.063 0.285 4.079 2.2

7

87 0.278 0.515 8.050 0.104 0.394 5.078 2.6

1 7 0.423 0.664 8.882 0.185 0.591 7.031 3.0

1 1

87 0.548 0.820 9.883 0.271 0.744 8.428 3.3

Source: American Concrete Institute (ACI).

the strong section is when the plywood exterior face grain runs parallel to the plywood

span (more common), and the weak section is when the plywood exterior face grain

runs perpendicular to the plywood span (not as common).

Stud andWaler Design Stud and waler sizes have many more options, so the rules

of material selection have more ranges. For instance, stud selection usually ranges

from 2 × 4 to 2 × 6 and 4 × 4. Double walers would include 2 × 4, 2 × 6, 2 × 8, 4 × 6,

and 4 × 8. Obviously, this still leaves a lot of decision making for the designer, but

considering the number of sizes available, this narrows down the field considerably.

This text provides the necessary properties of most available dimensional lumber

sizes for design purposes. Like plywood, rectangular dimensional has its weak

direction and its strong direction. This is usually very obvious because beams are

stronger when their depth is greater than their base. The tables will give both of

these values and the designer must be careful to use the correct size. As an example,

take a 2 × 4 and compare its section modulus about the x axis (strong) and the

y axis (weak).

S = bd2∕6

Strong∶ S = (1.5′′)(3.5′′)2

6= 3.063 in3

Weak∶ S = (3.5′′)(1.5′′)2

6= 1.313 in3

The strong direction in this case is 2.33 times stronger than the weak direction.

The only time the weak direction would be preferred is when the board is acting as

the sheeting as well as the stud, such as when footing and slabs are being formed.

Appendix 6 contains dimensional lumber properties for design purposes. Like

Table 9.4, this appendix has area, section modulus, moment of inertia, unit weight,

and the like. These values will be used for all dimensional lumber calculations.


This appendix also has a weak and strong section. The strong is using the materialin the vertical loading position (x-x axis) and the weak is loading the material in theflat, horizontal position (y-y axis).

Wood Form Design The process of form design will follow these steps:

1. Calculate the lateral pressure on the form.

2. Determine the plywood type and thickness.

3. Calculate the maximum stud clear spacing through shear, bending, and deflec-tion. Add the stud thickness to determine the on-center spacing.

4. Determine the stud size and calculate the uniform load on one stud using Pmax,the span, and the on-center spacing in step 3.

5. Calculate the waler spacing through shear, bending, and deflection.

6. Determine the waler size and calculate the uniform load on one double walerusing Pmax, the span, and the double waler thickness.

7. Determine the maximum load on the heaviest loaded tie.

Figure 9.4 shows typical wall forms using 2 × 4 and plywood construction.The following formulas will be usedwhen checking bending, shear, and deflection.

The span between components is the unknown variable in this case. In order to knowthe other variables for each check, the designer must know the material sizes andproperties. Figure 9.5 illustrates how multiple spans of plywood decrease bendingmoment and deflection.

FIGURE 9.4 Typical 2 × 4 and plywood formwork.


FIGURE 9.5 Plywood bending and deflection.

Bending

l = 10.95√FbS∕w

where l = span (in)

Fb = allowable bending stress of material (psi)

S = section modulus (in3)

W = uniform load on span (plf)

10.95 = constant

Shear—Plywood

L =Fv(lb∕Q)0.6w

where L = span (ft) divide by 12 to compare in inches

V = maximum shear (lb or k)

Fv = allowable shear stress

Ib∕Q = rolling shear constant (in2)

W = uniform load (plf)

0.6 = constant

Shear—Dimensional Lumber

L =F′v bd

0.9w+ 2d

12


F′v = allowable shear stress with shear factor of 2.0

b = base dimension of wood section (in)

d = depth dimension of wood section (in)

w = uniform load (plf)

0.9 = constant

Deflection

L = 1.69 3√EI∕w


where L = span (in)



E = modulus of elasticity

1.69 = constant

9.2.2 Form Design Examples Using All-Wood Materials

with Snap Ties or Coil Ties

Example 9.1 Cases 1 and 2will use amaximum lateral concrete pressure of 900 psf.

The calculations for form pressure were discussed at length in Chapter 8. It is assumed

that this value comes from a thorough pressure analysis, taking into account rate of

placement, temperature, mix design, wall height, and coefficient adjustments.

Case 1 will use the plywood in the weak direction, small studs, small walers, and

snap ties. This will represent a very lightweight system. The plywood direction is

intended to show the student how important it is to run the plywood in the strong

direction whenever possible. If not, it is either a mistake and the finished product will

indicate high deflection and bending failures; or, if it is intentional, the form design

will be less than economical as far as placement production and stud spacing.

Case 2 will use plywood in the strong direction, maximizing the plywood strength

and stud spacing, larger studs and walers, and higher capacity ties.

The example will alternate between case 1 and 2 so that we can see the differ-

ences between the two at each step. Also, all spacing calculations will be rounded to

the nearest tenth. In reality, these numbers will be rounded down to the nearest whole

inch, which makes sense with constructability practices and simplifying dimensions

for layout purposes. The fabrication of these forms must be done in an efficient man-

ner. Therefore, the easier (using dimensions that are quicker to read on a tapemeasure)

the spacing of the materials is, the lower the fabrication costs.

The material assumptions for cases 1 and 2 are recorded in Table 9.5.

TABLE 9.5 Case 1 versus Case 2 Assumptions

Case 1 Case 2

5

8

′′ BB-OES plywood used in the weak

direction

3

4

′′ BB-OES plywood used in the strong

direction

2 × 4 Douglas fir–larch studs 4 × 4 Douglas fir–larch studs

Double 2 × 4 Douglas fir–larch walers Double 3 × 8 Douglas fir–larch walers

5000-lb capacity snap ties She bolts

Step 1: Stud Spacing

Case 1: Plywood 5

8′′ thick BB-OES has been selected to be used in theweak direction.

The important aspect here is that the values used for this plywood must be the lower

values from Table 9.4, indicating that the plywood is not as strong when used in this

manner. Use the column labeled “weak direction.” For stress values use Table 9.3.


Case 1, Step 1: Check the maximum span between studs to comply with bending,

shear, and deflection requirements.

Bending

l = 10.95√(1600)(0.164)∕900 = 5.91 in

where l = span (in)

Fb = allowable bending stress BB-OES plywood in psi from Table 9.3

(1600 psi)

S = section modulus in in3 from Table 9.4 (0.164 in3)

W = 900 plf

10.95 = constant

The maximum span of the studs due to bending is 5.9 in when rounded to nearest

tenth.

Shear—Plywood

L = (60)(3.119)0.6(900)

× 12′′∕ft = 4.16′′


Fv = maximum allowable shear stess from Table 4.2 (60 psi)

Ib∕Q = rolling shear constant (in2) from Table 9.3 (3.119 in2)

W = 900 plf

0.6 = constant

The maximum span of the studs due to shear is 4.2 in when rounded to the nearest

tenth.

Deflection

l = 1.69 3√(1,500,000)(0.027)∕900 = 6.01′′

where l = span (in)

I = moment of inertia (in4) from Table 9.4 (0.027 in4)

W = 900 plf

E = modulus of elasticity from Table 9.3 (1,500,000 psi)

1.69 = constant

The maximum span of the studs due to deflection is 6.0 in when rounded to the

nearest tenth.

Step 1 is complete for case 1 and the maximum span between studs is the lowestvalue as determined by the three checks above. The lowest of the three checks is 4.2′′,which is governed by shear. From the clear span, the center-to-center spacing can be

determined by adding the stud thickness (1.5′′ for an S4S 2 × 4 on edge) to the span

(L) dimension.

Center to center = L + b = 4.2′′ + 1.5′′ (2 × 4) = 5.7′′ OC (on center)


Case 2: Plywood—3∕4′′ thick BB-OES is being used in the strong direction; so

the larger values are used from the tables, “strong direction.”

Case 2, Step 1: Check the maximum span between studs to comply with bending,

shear, and deflection requirements.

Bending

l = 10.95√(1600)(0.412)∕900 = 9.37′′

where L = span (in)

Fb = Allowable bending stress BB-OES plywood in psi from Table 9.3

(1600 psi)

S = section modulus in inches3 from Table 9.4 (0.412 in3)

W = 900 plf

10.95 = constant

The maximum span of the studs due to bending is 9.2 in when rounded to nearest

tenth.

Shear—Plywood

L = (60)(6.762)0.6(900)

× 12′′∕ft = 9.02′′


Fv = maximum allowable shear stress from Table 9.3 (60 psi)

Ib∕Q = rolling shear constant (in2) from Table 9.4 (6.762 in2)

W = 900 plf

0.6 = constant

The maximum span of the studs due to shear is 9.0 in when rounded to the nearest

tenth.

Deflection

l = 1.69 3√(1,500,000)(0.197)∕900 = 11.66′′

where l = span (in)

I = moment of inertia (in4) from Table 9.4 (0.129 in4)

W = 900 plf

E = modulus of elasticity from Table 9.3 (1,500,000 psi)

1.69 = constant

The maximum span of the studs due to deflection is 11.7 in when rounded to the

nearest tenth.

Step 1 is completed for case 2 and the maximum span between studs is the lowestvalue as determined by the three checks above. The lowest of the three checks is

9.0′′, which is governed by shear. From the clear span, the center-to-center spacing

can be determined by adding the stud thickness (3.5′′ for an S4S 4 × 4) to the span

(L) dimension.

Center to center = L + b, 9.0′′ + 3.5′′ (4 × 4) = 12.5′′ OC (on center)


Case 1 Stud: Determine the minimum spacing between walers using a 2 × 4

Douglas fir–Larch stud. The spacing of the walers is dependent on the stud loading

and size. From this information, the span between walers is determined. Since the

variables have been defined in detail in step 1, the remaining steps will not show the

formula in variable form (root formula).

Step 2: Waler Spacing

Case 1, step 2, span of walers. Use the same material in Table 9.3 as step 1. However,

use Appendix 6, which has been placed in Table 9.6, for the properties of the dimen-

sional lumber. The stud for case 1 is a 2 × 4.

TABLE 9.6 Properties of a 2 × 4 and 4 × 4 from Appendix 6

Material 2 × 4 4 × 4

Section modulus (in3) S4S 3.06 7.15

Base dimension (in) 1.5 3.5

Depth dimension (in) 3.5 3.5

Moment of inertia (in4) S4S 5.36 12.50

For Douglas fir–Larch construction grade:

Fb = 1000 psi

Fv = 95 psi × 2

for short-term use factor (shear is allowed twice the normal allowable shear stress by

ACI-347).

E = 1,500,000 psi

The uniform load (w) for each stud should be determined based on the spacing of

the stud and the Pmax concrete pressure. The new w for the case 1 2 × 4 stud is

W = 900 psf × (5.7′′∕12′′∕ft) = 427.5 plf

The uniform load is lower than the original Pmax because the spacing of the 2 × 4

studs is less than 1 ft.

Bending

l = 10.95√(1000)(3.06)∕427.5 = 29.3′′

The maximum span of the waler due to bending is 29.8′′ when rounded to the

nearest tenth.


There are two big differences in the shear calculation between plywood and dimen-

sional lumber. The first is the allowable shear being doubled for F′v when considering

a shear factor multiplier. The second is the addition of twice the depth (2d∕12) to the


span calculation. If these two allowances were not used, the span solution would be

extremely underestimated:

L =(F′

vbd)0.9w

+ 2d12

where L = span (ft) multiply by 12 to compare in inches

F′v = 95 psi × 2.0 = 190 psi

b = base dimension of wood section (1.50 in)

d = depth dimension of wood section (3.50 in)

w = uniform load (427.5)

0.9 = constant

L = 190(1.5)(3.5)0.9(427.5)

+(2 (3.5)12

)= 3.173 × 12′′∕ft = 38.08 in

The maximum span of the waler due to shear is 38.1′′ when rounded to the nearesttenth.

Deflection

l = 1.69 3√(1,500,000)(5.36)∕427.5 = 44.94′′

The maximum span of the waler due to deflection is 44.9′′ when rounded to the

nearest tenth.

Step 2 is completed for case 1 and the maximum span between walers is the lowestvalue as determined by the three checks above. The lowest of the three checks is

29.3′′, which is governed by bending.Case 2, Step 2: Determine the waler spacing for case 2 using a 4 × 4 stud.

The uniform load (W) for each stud should be determined based on the spacing of

the stud and the Pmax maximum concrete pressure. The new W for the 4 × 4 stud is

W = 900 psf × (12.5′′∕12′′∕ft) = 938 plf

The uniform load is greater than the original Pmax because the spacing of the 4 × 4

studs is slightly greater than 1 ft on center.

Bending

l = 10.95√(1000)(7.15)∕938 = 30.23 in

The maximum span of the waler due to bending is 30.2′′ when rounded to the

nearest tenth.


L = 190(3.5)(3.5)0.9(938)

+[2 (3.5)12

]= 3.34′ × 12′′ = 40.08′′

The maximum span of the waler due to shear is 41.2′′ when rounded to the nearesttenth.


Deflection

l = 1.69 3√(1,500,000)(12.50)∕938 = 45.87′′

The maximum span of the waler due to deflection is 45.9′′ when rounded to the

nearest tenth.


30.2′′, which is governed by bending.

Step 3: Tie Spacing

The difference between steps 2 and 3 is the fact that the waler is a double member

so all the section properties (S, I, d, and b) are doubled. The waler is a double beam;

therefore, it is twice as strong. Table 9.7 shows values for the waler material from

Appendix 6.

TABLE 9.7 Properties of a 2 × 4 and 3 × 8 from Appendix 6

Double Waler Material Double 2 × 4 Double 3 × 8

Section modulus (in3) S4S 3.06 × 2 ea 21.9 × 2 ea

Base dimension (in) 1.5 × 2 ea 2.5 × 2 ea

Depth dimension (in) 3.5 × 2 ea 7.25 × 2 ea

Moment of Inertia (in4) S4S 5.36 × 2 ea 79.39 × 2 ea

Before the spans can be calculated, the uniform load on the double waler must

be determined. The tie spacing is dependent on the waler size, waler type, and the

uniform load on each waler.

Case 1Waler—Determine the minimum spacing between ties using a double 2 × 4

Douglas fir–Larch waler. The spacing of the ties is dependent on the waler loading

and size. From this information, the span between ties is determined. Once again,

since the variables have been defined in detail in step 1, the remaining steps will not

show the root formulas.

Case 1, Step 3, Span of Walers (tie spacing)—Use the same material (Table 9.2)

as step 1. However, use Appendix 6 for the properties of the dimensional lumber. The

waler for case 1 is a 2 × 4.

The grade of lumber has not changed from the studs to the double walers. For

Douglas fir–Larch construction grade:

Fb = 1000 psi

Fv = 95 psi × 2 = 190 psi

for short term use factor (shear is allowed twice the normal allowable shear stress by

ACI-347).

E = 1,500,000 psi


The uniform load (W) for each waler should be determined based on the spacing

of the waler and the Pmax concrete pressure. The newW for the case 1, 2 × 4 waler is

W = 900 psf × (29.39′′∕12′′∕ft) = 2205 plf

The uniform load is greater than the original Pmax because the spacing of the 2 × 4

double waler is more than 1 ft.

Bending

l = 10.95√(1000)(3.06)(2)∕2205 = 18.2 in

The maximum span of the waler due to bending is 18.2 in when rounded to the

nearest tenth.


L = 190(1.5)(3.5)(2)0.9(2205)

+[2 (3.5)12

]= 1.593′ × 12′′ = 19.1 in

Themaximum span of the waler due to shear is 19.1 in when rounded to the nearest

tenth.

Deflection

l = 1.69 3√(1,500,000)(5.36)(2)∕2205 = 32.8′′

The maximum span of the waler due to deflection is 32.8 in when rounded to the

nearest tenth.


18.2′′, which is governed by bending.Case 1 formwork is represented by Figure 9.6 where the forms are made from

2 × 4’s and plywood except the plywoodwas used in the strong direction in this photo.

Case 2, Step 3: Determine the tie spacing for case 2 using a 3 × 8 double waler.

The uniform load (W) for each waler should be determined based on the spacing

of the waler and the Pmax concrete pressure. The new W for the case 2, double 3 × 8

waler is

W = 900 psf × (30.2′′∕12′′∕ft) = 2265 plf

Bending

l = 10.95√(1000)(21.9)(2)∕2265 = 48.15′′

The maximum span of the waler due to bending is 48.2 in when rounded to the

nearest tenth.


L = 190(2.5)(7.25)(2)0.9(2265)

+[2 (7.25)

12

]= 4.59′ × 12′′ = 55.05′′


FIGURE 9.6 Junction box formed with all-wood forms.

Themaximum span of the waler due to shear is 54.4 in when rounded to the nearest

tenth.

Deflection

l = 1.69 3√(1,500,000)(79.39)(2)∕2265 = 79.76′′

The maximum span of the waler due to deflection is 79.8 in when rounded to the

nearest tenth.


48.2′′, which is governed by bending.

Step 4: Tie Capacity

The tie capacity and size needs to be selected to complete this wood form design. The

load on an individual ties is

Ptie = [tie spacing (ft)] × [waler spacing (ft)] × Pmax

Case 1∶ Ptie = [18.2∕12′′∕ft] × [29.3∕12′′∕ft] × 900 psf = 3333 lb in tension

A 3500-lb tie will work for this case.

Case 2∶ Ptie = [48.2 ∕12′′∕ft] × [30.2∕12′′∕ft] × 900 psf = 9098 lb


A 1∕2′′ × 3∕4′′ Shebolt will work for this case, capacity (safe working load includes1.5 : 1.0 FOS) = 12,000 lb.

Solutions to wood form design example 1, cases 1 and 2, are shown in Table 9.8.

TABLE 9.8 Case 1 vs. Case 2 Solutionsa

Material

Case 1 Center-to-Center

Spacing (in)

Case 2 Center-to-Center

Spacing (in)

Plywood5

8W

3

4S

Stud spacing 5.7 12.5

Waler spacing 29.3 30.2

Tie spacing 18.1 48.2

Tie type and load Snap tie (3333 lb) She bolt (9098 lb)

aW = weak and S = strong.

9.2.3 Formwork Charts

Before the next example is illustrated, the use of wood formwork charts will needto be mastered. Such charts are very convenient because they eliminate much of the

tedious calculation performed in the earlier examples, thus avoiding the risk of errors.These charts, based on NDS 2001, can be found in Appendix 7 and are provided withpermission from Williams Form Engineering Corporation. The following items areneeded to use the charts for wood formwork design:

Allowable bending stress of wood specie or plywood

Allowable shear or rolling shear stress of wood specie or plywood

Modulus of elasticity of wood specie or plywood

Uniform load (plf) on supporting beam

Number of spans for the supporting beam (single, 2, 3, or more)

Plywood and beam sizes

For typical all-wood form design, these three charts are necessary:Plywood Chart—This includes two main sections, stronger plywood (higher

allowable stress) and weaker plywood (lower allowable stress), each with twocolumns: one for plywood used in the strong direction and one in the weak.

Single-Beam Chart—This is used for studs, joists, and other single-beam com-

ponents. The number of spans is an option because the chart will use either wl2∕8 orwl2∕10.

Double Walers—This is used for any two-beam component (double walers), forwhich all the applicable beam properties have been doubled to take into account the

two beams working together.The fine print on these charts indicates that the dimensions shown are center-

to-center dimensions and not clear span, as long as the beam has a “relatively narrow”

width (b) dimension. This author will loosely interpret this to mean a 2× beam up toroughly a 4× beam.


The form design process using these charts follows:

Determine maximum lateral pressure from the concrete (see Chapter 8).

Use the applicable plywood chart and determine the stud spacing.

Calculate the uniform load on one stud based on its spacing.

Use the applicable single beam chart and determine the double waler spacing.

Calculate the uniform load on a double waler based on its spacing.

Use the applicable double waler chart and determine the tie spacing.

Calculate the load on the heaviest loaded tie. Typically, these are the ties in thesecond row from the bottom of the form.

Example 9.2 Formwork Design Using Appendix 7 Charts (courtesy of WilliamsForm Engineering Corporation).Design a wall form to meet the following conditions:

Wall form 12 ft high × 16 ft wide

R = 5 ft∕hT = 65∘F

Materials:Plywood 3∕4′′:

Allowable Stress Values:

Fb = 1930 psi

Rolling shear = 72 psi

E = 1,500,000 psi

Deflectionl

360, but not to exceed

1

16′′

Dimensional Lumber:

2 × 4 studs

Double 2 × 6 walers

Snap or taper ties

Allowable Stress Values:

Fb = 975–1500 psi

F′v = 180 psi; F′

v = Fv × 2

E = 1,600,000 psi

Deflectionl

360, but not to exceed

1

4′′

Step 1: Calculate Pmax.

Use formula 1 from Chapter 8 since h < 14 ft and R < 7 ft∕h

Pmax = 150 + (9000 × 5)65

= 842 psf

See Figure 9.7 for a diagram of the resulting pressure.


FIGURE 9.7 Pressure diagram.

Step 2: Determine stud spacing using 3∕4′′ plywood.From Appendix 7, 842 psf,Fb = 1930 psi, shear = 72 psi

Space studs at 12′′ OC

Step 3: Determine waler spacing with a 2 × 4 stud at 12′′ OC.

W = 842 psf × 12′′∕12′′∕ft = 842 plf

Stud and joist beam table in Appendix 7, 842 plf, fb = 1500 psi, H =180 psi

Space double walers at 28′′ OC. Use 28′′ OC.

Step 4: Determine tie spacing with 2 × 6 double walers at 25′′ OC.

W = 842 psf × 28′′∕12′′∕ft = 1965 plf

Double waler table in Appendix 7, 1965 plf, fb = 1250 psi,H = 180 psi

Space ties at 37′′ OC. Use 36′′ OC.

The tables in Appendix 7 give you “not-to-exceed” numbers, center to center, so

they should be adjusted to work with the panel size and to make the layout for the

carpenters straightforward. Therefore, use the values that have been rounded down.

Tie load = 842 psf × 28′′∕12′′∕ft × 36′′∕12′′∕ft = 5894 lb

Final design:

3∕4′′ plywood (used in the strong direction)

2 × 4 studs at 12′′ OC

2 × 6 double walers at 28′′ OC

Ties at 36′′ OC with minimum SWL ≥ 5894-lb capacity


FIGURE 9.8 Composite form using aluma beams and steel channels.

Formwork Design Using Lumber, Aluminum Beams, and Double Steel

C-Channels All wood forms are versatile and inexpensive. However, when a con-tractor desires a larger form, with materials that are more reusable, they might con-sider using a combination of plywood, dimensional lumber, or aluminum beams forstuds and aluminum or steel double walers. This next section will use these materi-als for form design Examples 9.3 and 9.4. Figure 9.8 is an example of a larger type,composite form design. Most of the materials on this form are reusable from projectto project.

Example 9.3 Composite Form DesignUsing a lateral concrete pressure of 981 psf (which is assumed to be precalculated

using the methods from Chapter 8), design a two-sided wall form using 3∕4′′ BB-OESplywood in the strong direction, 4 × 4’s for horizontal studs, double steel American

C-Channels for vertical walers and she bolts. Do not exceed a1

2′′ diameter inner rod

she bolt. The following data should be used for the design criteria. Ignore deflectionin all calculations.Material Properties:Plywood:

Fb = 1600 psi

Fv = 60 psi

Dimensional Lumber—Hem fir construction grade lumber

Fb = 975 psi


Fv = 75 psi,F′v = 150 psi

E = 1,300,000 psi

A36 steel with a Fb = 0.6;Fy = 21,600 psi

Step 1: Determine the safe spacing of the 4 × 4 horizontal studs by analyzing a

12′′ strip of plywood. Also, check bending and shear, and ignore deflection. Do not

use Appendix 7.

W = 981 psf × 1′ = 981 plf

Bending—Plywood

l = 10.95√(1600)(0.412)∕981 = 8.8 in

Shear—Plywood

L = (60)(6.762)(0.6) (981)

× 12′′∕ft = 7.9 in (governs)

Space the 4 × 4’s at 7.9′′ + 3.5′′ (4 × 4 width) = 11.4′′ use 11′′ OCStep 2: Determine the safe spacing of the double steel channels. In other words,

how far apart can the double steel channels be spaced so the 4 × 4’s do not fail in

bending or shear.

W = 981 psf × 11′′∕12′′∕ft = 899 psf

Bending—Studs

l = 10.95√(975)(7.15)∕899 = 30.5 in (governs)

Shear—Studs

L =[150 (3.5) (3.5)

0.9(899)

]+[2 (3.5)12

]= 2.85′ × 12′′∕ft = 34.2 in

Space the double steel channel’s at 30.5′′ OC.Step 3: Determine the double waler size and determine the tie capacity. At this

point in the design, experience can come in handy because tie spacing can be esti-

mated and the double channel can be sized appropriately, or the channel size can be

known and the maximum tie spacing can be calculated. For this example, the ties

will be spaced to stay within the 1∕2′′ diameter requirement stated above and then the

double channel will be sized.

A1

2′′ inner rod tie for a she bolt can safely support 12,000 lb (1.5 : 1.0 FOS) accord-

ing to Williams Form Engineering Corporation. If the double walers are at 34′′ OC,then the tie spacing can be estimated.

Pcapacity = Area × 981 psf; area = tie spacing × double waler spacing, therefore:

12,000 lb = (34′′∕12′′∕ft) × (tie spacing) × 981 psf

Tie spacing = 4.32 ft say 4.25 ft


The uniform load on the double waler isW = 981 psf × 34∕12′′∕ft = 2776 plf.

W = 2776 plf

Figure 9.9 is an example of the FDB for this double waler.

FIGURE 9.9 FBD of double waler.

Waler size is based on the maximum moment generated by the FBD diagram in

Figure 9.9. A computer program could be used, or the maximum moment can be

calculated withM = wL2∕10.

M for a continuous beam over multiple supports (ties)

M = 2776 plf (4.25′)2∕10 = 5014 ft-lb

Smin = M∕Fb, S = (5014 ft-lb)(12′′∕ft)∕21,600 psi = 2.79 in3

Since a double steel channel waler is a “double” beam, the minimum section mod-

ulus can be divided by 2 before searching for the best channel:

Smin for a single channel = 2.79 in3∕2 = 1.395 in3

Some channels that are in the range of this requirement are shown in Table 9.9.

The best is shown in bold.

TABLE 9.9 Steel Channel Properties

Channel Size Section Modulus (Sx) Weight per foot Conclusion

C 𝟒 × 𝟓.𝟒 1.93 5.4 BESTC 3 × 6 1.38 6 Section too small

C 4 × 7.5 2.29 7.5 OK, but heavy

The conclusion for this channel selection should be that the C4 × 5.4 is best for

strength and economy. The C3 × 6 is very close; however, it is heavier than the chan-

nel chosen and it is also weaker.

Final design:

3∕4′′ plywood (strong)

4 × 4 studs horizontal at 11′′ OC

Double C4 × 5.4 channel walers at 34′′ OC1∕2′′ she bolts at 4.25 ft OC, tie load not to exceed 12,000 lb

The tie lengths would have to be calculated so as to order the correct equipment.

Figure 9.10 illustrates the hardware required for this design.


FIGURE 9.10 She-bolt cross section.

Here is how that would be accomplished for a she-bolt system:

Assume the wall thickness is 16′′ reinforced concrete.

Available lengths for 1∕2′′ × 3∕4′′ she bolts are 20′′, 24′′, and 28′′.

TABLE 9.10 She-Bolt Length Calculation

Item Length (in)

Plywood × 13

4

4 × 4 studs × 1 3.5

Double C-Channel × 1 4

Plate washer × 2 3

Wing nut × 2 1.5

Excess threads × 2 3

Required she bolt L 15.75

Add 2′′ of taper inside wall 2

Net system L 17.75

The minimum thread length for this size tie is 9′′. Therefore, there is plenty of

thread at the end on the tie so that the wing nut will not “bottom out” (run out of

thread).

Use the 1∕2 × 3∕4 she bolt × 20′′ in length.Finally, the coil rod would be cut the wall thickness less 1-in cover from each face

of wall, 16′′ − (2 × 1′′) = 14′′.

Aluminum Beams Aluma beams are extruded aluminum beams that (1) are shapedfor maximum bending and deflection capacity, (2) provide a slot for a “nailer”, and(3) allow for a slotted bolt to attach walers.

For these reasons, the beam is shaped as shown in Appendix 9. The pocket in thetop is for a wood or fiberglass 2 × 2 that allows for a sheet of plywood to be nailed orscrewed to it. The slot in the bottom is for a slotted bolt that attaches an aluminum,steel, or wood waler. Appendix 9 also includes design information for aluminumbeams to be used for form design.


aluminum beam design with manufacturers design charts On many occa-sions the manufacturer of a particular component provides the engineering data

necessary to use its product. This data is easily obtained by contacting the localrepresentative of the company that provides such a product. The data received willnot always follow a typical protocol. For instance, the information in Appendix 9(courtesy of Aluma Systems) is from an aluminum beam manufacturer. This chartutilizes maximum uniform load information based on some spacing shown on the

chart. The product manufacturers have precalculated bending, shear, and deflectionto determine the maximum allowable uniform load.

Before moving on to Example 9.4, the aluminum beam chart should be studied.The chart is broken up into an x and y axes. The x axis represents the type of span:simple, two spans, or three spans. Most of the time formwork components span three

or more spacings. The y axis represents the span of the aluminum beam betweensupports, from 4 to 12 ft. The values within the table in Appendix 9 are the allowableuniform loads on a single beam at the given span. The fine print of the chart has someuseful information. The letters designate which of these three criteria governed thespacing for the load:

R denotes that shear governs

M denotes that bending moment governs

D denotes that deflection governs using L∕360 criteria

* denotes deflection governs not exceeding1

4′′ deflection

FOS = 2.2 : 1.0

The first step to using this type of chart is to determine the spacing of the aluminum

beams, which is based on the loading and the plywood thickness. Once the uniformload is known, the number of spans is estimated (single, two, three, or more). Now,the span length of the aluminum beam can be found on the y axis. This is the farthestthe beam can span given the other conditions.

Example 9.4 Composite Form DesignUsing a design pressure of 1100 psf and the wood design charts from Example 9.2,

design a two-sided wall form using 3∕4′′ HDO plywood in the strong direction (see

Table 9.11), aluminum beams (Appendix 9) and steel double channels (Appendix 1).

TABLE 9.11 High-Density Overlay (HDO) Plywood Propertiesa

Stud Spacing (in)5

8

′′ (plf)3

4

′′ (plf) 11

8

′′ (plf)

8 1970 2050 2230

9 1600 1800 1976

10 1200 1500 1660

12 745 1060 1380

16 350 505 1000

19.2 125 305 555

aHDO plywood used in the strong direction and complying with L/360

deflection limits.


Do not exceed the capacity of a 11

8′′ to 7

8′′ diameter C7T taper tie (SWL = 24,000 lb).

Assume one wall form measures 20′ high × 24′ wide.Figure 9.11 shows the allowable loads for HDO plywood in graph form.

FIGURE 9.11 Allowable load graph for HDO plywood (Courtesy of Olympic Plywood).

Step 1: The concrete lateral pressure has been given at 1100 psf.

Step 2: Determine stud spacing using 3

4′′ HDO plywood.

Using the design load of 1100 psf and Table 9.12 for3

4′′ HDO plywood, determine

the stud spacing. The table indicates that the studs can be spaced at 10′′ OC:Step 3: Determine double waler spacing with an aluminum stud at 10′′ OC:

W = 1100 psf × 10′′∕12′′∕ft = 913 plf

The aluminum beam chart (Appendix 9) is now used to determine the spacing

of the double walers. Aluminum beams are very strong and many times the spacing

given is more than the rest of the form components can support. Using the column

in Appendix 9 titled “3 Span (lbs./ft.)” and based on 913 plf, an aluminum beam

can span 8.0 ft if L/360 is required and 7.5 ft if deflection is limited to1

4′′. It will be

assumed that the more stringent requirement is necessary; therefore, space double

steel walers at 7.5 ft on center.

As mentioned above, this spacing will increase the loads on the waler and the

ties. The waler has to be designed with a 7.5-ft tributary width or W = 1100 psf ×7.5 ft = 8250 plf waler linear load. It will soon be determined whether or not this

load makes the system economical. A large part of formwork costs are in the tie

placement and removal. Therefore, if the tie quantity can be reduced, costs can typi-

cally be lowered. However, if this reduction in quantity results in a very large tie, the

objective could be lost. This example will be completed taking these thoughts into

consideration.

If the walers are spaced at 7.5 ft on center,W = 8250 plf.


Step 4: Determine what the tie spacing would be using the prescribed 24,000-lb

SWL taper tie.

Maximum tie spacing = 24,000-lb tie∕8250 plf = 2.91 ft

If the ties for a 20-ft-tall wall formwere spaced at 2.9 ft on center, then there would

be approximately 7 rows of ties × 4 walers = 28 ties. If a 5 × 5 tie and waler pattern

was employed, for instance, then the form would require approximately 20 ties, and

the tie spacing would increase to between 4 and 5 ft. To do this, however, the walers

would also be spaced between 4 and 5 ft and the aluminum beams would not be used

to their maximum capabilities. However, this is usually the correct option because

one additional waler is more economical than 8 more ties. It should be pointed out

that besides having to install and remove the ties, the ties have to be prepped and the

holes left by the ties have to be patched by a mason/finisher. After considering all the

labor that is associated with each tie (3 to 6 man-hours per tie, $200–$500 each in

labor and materials combined), it becomes obvious that it is important to reduce the

quantity whenever possible.

Another approach would be

24,000-lb tie

1100 psf= 21.82 SF per tie

This creates an average combination tie and waler spacing of 4′8′′ by a tie spacing of4′8′′ or other combinations that equal approximately 22 SF.

Taking all of this information into consideration and spacing the walers at 5′0′′ oncenter and the ties at 4′6′′ on center produces a layout of 6 walers with 20 ties. This

layout is illustrated in Figure 9.12. One can then compare the cost between using 2

more walers versus 8 more ties.

FIGURE 9.12 Panel layout for Example 9.4.

For two additional walers, purchase, install, and remove walers:

Material∶ 21 ft × 20 lb∕ft (estimate) × 2 ea = 840 lb of steel at $0.60∕lb =$546


Labor∶ 2 ea at 2 man-hours∕each (estimate) = 4 man-hours × $50 = $200

Total∶ $746 for two additional walers

For additional ties, prepare, install, remove, and patch hole: Using the price range

above and an average of $350 per tie, eight additional ties would cost 8 × $3500 =$2800:

$2800 – $746 = $2054 savings by using the 6-waler, 20-tie design

Also take into account that the walers are placed one time during form fabrication,

whereas the eight ties repeat as many times as the form is reused. If the form is used

eight times, for example, the savings becomes over $21,000.

Step 5: Design the double waler using the waler and tie spacing above.

W = 1100 psf × 5′ = 5500 plf

Figure 9.13 shows a loading diagram of a double channel.

Mmax = 5500 plf(4.5′)2∕10 = 11,138 ft-lb

Smin = (M∕Fb)∕2 channelsSmin = [(11,138 ft-lb (12′′∕ft)∕21,600 psi]∕2 channels = 3.09 in3

FIGURE 9.13 Loading diagram of a double channel.

Using American Standard Channels, the choices close to this minimum section

modulus are listed in Table 9.12. The C6×8.2 appears to be the best of the three.

TABLE 9.12 Available Channel Sizes for Example 9.4

Channel Size Section Modulus (Sx) Weight per Foot Conclusion

C5×6.7 3.0 6.7 Section too small

C5×9 3.56 9 OK, but heavy

C6×8.2 4.38 8.2 OK, lightest, BEST

The C6×8.2 also has a moment of inertia of 13.1 in4. This would also help in

minimizing the deflection in the double waler because it also has a higher moment of

inertia, which would lower the maximum deflection. The C5×6.7 channel could be

used if the tie spacing was reduced a few inches. Contrary to that, since the C8×6.2channel is overdesigned, the tie spacing also could be increased to maximize the

waler size. However, this would contradict the goal to not exceed a 24,000-lb taper tie.

All these points should be examined to make sure the design is the most economical.


Final Design Conclusion:

3∕4′′ HDO plywood

Aluminum beam studs spaced at 10′′ on center

Double C6×8.2 American Steel Channels spaced at 5 ft on center

11

8′′ × 7

8′′ C7T taper tie (SWL = 24,000 lb) spaced at 4.5 ft OC

Figure 9.14 shows a taper tie being used with an all-wood form system. The tie

lengths would have to be calculated to order the correct equipment. Here is how that

would be accomplished for a taper tie system.

FIGURE 9.14 Taper tie cross section.

Assume the wall thickness is 16′′ reinforced concrete. Table 9.13 lists the mea-

surements necessary for calculating the length of the she bolt.

TABLE 9.13 She bolt Length Calculation

Item Length (in)

Concrete 16

Plywood × 2 1.5

Aluminum beams × 2 13

Double C − channel × 2 12

Plate washer × 2 4

Wing nut × 2 3

Excess threads × 2 6

Required taper tie L 55.5

Available taper tie lengths are 30′′, 34′′, 36′′, 42′′, 48′′, 54′′, and 60′′.The 54′′ tie would work, and the excess thread availability would be reduced from

6′′ to 4.5′′.The 60′′ tie will also work if the thread lengths are long enough that the wing nut

does not “bottom out” (run out of threads). This tie has 10′′ of thread at the long endand 8′′ of thread at the short end. Based on this data, the wing nut would not run outof threads.


CASE STUDY: RADIUS FORMWORK, PLEASANT GROVE

WWTP CONSTRUCTION

Project Overview

Waste water and water treatment plants are great projects to study in construc-

tion because they involve almost every type of construction including exca-

vation, mechanical, electrical, underground, concrete structures, pile driving,

and all aspects of architecture. The Pleasant GroveWastewater Treatment Plant

(WWTP) in Roseville, California, was no exception. This project was a brand

new plant for the City of Roseville, making it the city’s second WWTP. The

growth of the city required increased capacity, but the original plant had already

been expanded and was running out of real estate. Also, the new plant location

was closer to the city’s expanding areas thus added convenience.

A large portion of this project involved new concrete structures to transfer

and treat water throughout the hydraulic flow process. These structures ranged

in size and shape, from large rectangular structures to small, round, or tall struc-

tures. The focus of this case study is on three structure types that had curved,

vertical walls. The first type included four clarifier tanks that had a 65-ft average

radius and stood 16 ft tall. The second type involved three oval-shaped oxida-

tion ditches that had curved ends measuring 60 ft in diameter and stood 16.5 ft

tall. The third type included two digester tanks that had a 25-ft radius walls and

stood 32 ft tall.

It was decided early in the form decision process that the digester, with a

much different radius and wall height, would have a separate form system than

the other two. The 32-ft height and the fact that there were only eight wall place-

ments (four 90∘ pours per digester) directed the decision to rent a steel form.

The project looked at EFCO and SYMONS. The two systems are very different.

The EFCO system uses the “ready radius” panels that can be curved to a desired

radius on site. The SYMONS system comes with precurved walers that are cus-

tom fabricated for the desired radius. After a cost analysis, the project decided

to go with the SYMONS form (shown in Figure 9.CS1).

Like many decisions in construction, they are made with very detailed cost

analyses, which include estimated production rates. However, since only one

method is eventually used, the method not chosen is never tested. Therefore, a

final, actual cost comparison can never be made so the contractor must accept

the decision. At this point, all efforts should be focused on making the chosen

option as cost-effective as possible.

Since the digester wall decision was made, now the clarifiers and oxidation

ditch walls needed to be designed. The geometry between the two structures

were almost identical and the number of wall placements made any decision

potentially a big “money maker” or a big failure. All in all, there would be a

(continued)


CASE STUDY: RADIUS FORMWORK, PLEASANT GROVE

WWTP CONSTRUCTION (Continued)

FIGURE 9.CS1 Steel forms used for a digester tank.

total of 44 wall placements of walls with 60- to 65-ft radii and 16- to 16.5-ft

height. The similarities in these two structures had to be taken advantage of. The

wall height was easy. The form engineers knew that a 16′9′′ panel would workwith all 44 walls. It was the radius that was in question. Could a form panel

be made that could be flexible enough to work with both the 60-ft radius and

the 65-ft radius? Could a form be made with a 62.5-ft radius that would work

for all 44 wall placements? The answer was yes. An aluminum vertical studded

panel was made with horizontally curved double steel channels (for the 62.5-ft

radius) that could open up to a 65-ft radius and close up to a 60-ft radius. The 3

4′′

BB-OES plywood had no problem making the radius and no provisions were

necessary to weaken or wet the plywood for flexibility. Figure 9.CS2 shows

this form being used on a clarifier wall.

This form is a custom-job-built radius form using plywood, aluminum

beams, and rolled double steel channel walers. The same process used in

Example 9.4 would be used to design this form. One difference would be that

the plywood would be changed to BB-OES or MDO because the HDO would

be difficult to bend into the necessary radius. Another difference would be that

the aluminum studs would be turned in the vertical direction and the double

steel channels would be placed horizontal and curved to the radius of the

permanent structure geometry. These forms were lightweight, approximately

15 psf, and were built for the tanks that did not exceed 16.75 ft in height.


FIGURE 9.CS2 Aluminum and rolled steel channel radius forms.

Formwork Geometry Form design is not just about analyzing the forces from the

concrete and stresses of the material. Structures have all sorts of shapes and sizes.

Sometimes the more difficult task is designing forms that conform to the shape of

the permanent structure. Many times, features have to be added in order to solve a

problem that the geometry caused.

Corners, curves, and intersections are all characteristics of concrete structures that

add difficulty to the design process. It is not always the fact that the wall is tall and

the pressure from the concrete is great. Sometimes a small structure can have none

of these characteristics except that the corners and intersections of wall create weak

spots in the form design. At times, the form designer spends more time figuring out

how one form panel is going to attach to another form panel, or how to tie a corner

or intersection together when the ties do not have a proper termination point on the

opposite side of the wall. Figure 9.15 illustrates this point for both a corner form con-

nection and an intersection. The arrows represent unsupported sections (weak points)

in the forms.

form drawings Once the calculations for a design are complete, the work has just

begun, so to speak. Someone has to take the calculations, which are typically not-to-

exceed values, and create the panel and layout drawings. These drawings show plan

views and sections of the elements and also refer to accompanying individual panel

drawings that detail the panel for size, materials, spacing of members, and tie loca-

tions and quantities at a minimum.


FIGURE 9.15 Formwork corners and intersections.

CASE STUDY: SOUND WALL CONSTRUCTION OVER

BIDWELL PARK, CHICO, CALIFORNIA

Project Overview

When Highway 99 was widened over prestigious Bidwell Park in Chico,

California, between 2011 and 2014, the contractor was faced with many

challenges. Working in or over Bidwell Park brings the attraction of many

groups and individuals that do not support change to the park. When John and

Annie Bidwell donated this land to the City of Chico, the agreement was very

specific to leave the park as is whenever possible. Years of planning, permits,

and negotiations preceded the construction of the Highway 99 project.

Another challenge facing the contractor was protecting Big Chico Creek

(shown in Figure 9.CS3), which runs in the middle of the narrow 3600+-acrepark and under the Highway 99 overcrossing and eventually through downtown

Chico and Chico State University.

The new piers for the widened bridge were designed to be constructed adja-

cent and sometimes within the creek low-water line. In the third year of the

project, just when one thought all the challenges were behind the contractor,

a final logistic/access issue had to be planned. The new sound wall, which is

attached to the bridge barrier rail, is 30 ft above the ground on one side and

is adjacent to two lanes of live traffic on the other side. The work access lane

available to the contractor was less than 20 ft in width. A small hydraulic rough

terrain crane could barely fit within these limits and would not allow any other

access when being used. In addition, an operator would have to be on the crew

continuously for the duration of the operation.

Furthermore, if a crane were placed on the ground level, the access

along the park and the city streets would have made it very difficult, if not

impossible, to set up a crane that could reach the portions of the wall that it

needed to reach. Figure 9.CS4 illustrates the wall and the basic geometry of

the structure.


FIGURE 9.CS3 Big Chico Creek.

FIGURE 9.CS4 Schematic of sound wall.

Traveler

The contractor had to figure how to form the cast-in-place, two-sided wall from

the bridge with just the right lane and shoulder closed. The contractor employed

an EFCO form system and developed a traveling truss that rolled along the

closed lane and supported both the inside and outside forms. Figure 9.CS5A

shows the truss beam that is supporting the inside and outside forms.

Once the wall forms were placed on the truss beam, they never had to be

hoisted with a crane again. From this point forward, the truss, which was

mounted on casters, was moved with a forklift. The wall panel taper ties were

(continued)



BIDWELL PARK, CHICO, CALIFORNIA (Continued)

FIGURE 9.CS5A Inside and outside forms supported by cantilevered beam.

FIGURE 9.CS5B Crane hoisting first form into place.


removed, the panels were pulled away from the wall and the truss was pushed

40–50 ft to its next wall section. Figure 9.CS5B shows the first wall placement

from a section that the crane could access from the ground level.

Truss and Counterweights

EFCO designed a truss and counterweight system shown in Figure 9.CS6. The

truss was made from EFCO equipment and components. The wall forms were

their standard wall panel using plywood, E-beams, and EFCO double steel

channels (super studs).

FIGURE 9.CS6 Traveling truss with counterweights.

The counterweights were precast onsite to custom dimensions and weights

that fit the framework of the truss and kept the forms supported in position

(see Figure 9.CS7).

Form Fabrication with Form Liner

In addition to the access complications, the wall design required form liner

that cast a scenic imprint into the wall for aesthetic purposes. The contractor

mounted a polyurethane form liner that usually costs between $30 and $60 per

SF onto the form and repeated the pattern with each form use. Figure 9.CS8

shows the wall forms being fabricated with the form liner prior to the first

placement.

(continued)



BIDWELL PARK, CHICO, CALIFORNIA (Continued)

FIGURE 9.CS7 Counterweights (deadmen) on casting bed.

FIGURE 9.CS8 Prefabricated wall forms with form liner.


9.2.4 Estimating Concrete Formwork

Using Form Ratios to Accurately Verify a Concrete Estimate Reinforced

concrete construction estimating requires a great deal of labor and temporary struc-

ture knowledge. When the finished product is delivered, one may not realize how

much work went into the creation of the structure. Competently estimating this work

involves a great deal of past project experience, and form ratios created by experts

can be the link that brings past experience to the estimator.

Reinforced concrete estimating utilizes several categories of costs. Labor is one

of if not the largest cost element. The formwork, the reinforcing steel, the place-

ment of the concrete, the finishes, and some additional miscellaneous items are the

other cost categories. The furnishing and placing of reinforcing steel is traditionally

a subcontractor cost. Materials are also required in two categories: permanent and

temporary. STS (small tools and services or supplies) is another cost category. Finally

the cost of construction equipment, such as forklifts, cranes, and pickup trucks, must

be included.

For consistency, examples shown in this book will use labor rates at $40.00 per

labor hour. Other costs shown are actual project costs from the San Francisco Bay

Area and the Central Sacramento Valley (northern California) that have been altered

slightly to preserve confidentiality.

What Is a Form Ratio? A form ratio (FR) is developed by dividing the neat volume

of concrete in cubic yards (CY) for a structural element into the total contact area of

formwork in square feet of the same element as shown below:

FR = total forms (SFCA)∕total volume (CY)

The formwork quantity should include any concrete surface that requires a form

to support the plastic concrete during its curing process. This should include footing

and slab-on-grade edge (SOG) forms, all wall surfaces, the formed area of all elevated

concrete surfaces (excluding additional surfaces for access platforms), construction

joints, miscellaneous curbs and equipment pads, and all other formed surfaces. The

volume of concrete should include all concrete elements except fill concrete and other

concrete not requiring a form. However, unformed concrete volume is still accounted

for in the estimate because the unformed concrete must be purchased and there may

be some incidental costs associated with this volume in the estimate. During actual

construction, it is crucial that the project staff record actual costs in this same manner.

This will be discussed more when we discuss the historical cost data.

Table 9.14 illustrates that the larger the element’s thickness, the smaller the ratio.

Also, if an element is supported by the ground (such as footings and SOGs), the ratio

is lower because no formwork is required beneath the concrete. Wall concrete requir-

ing forms on both sides generates higher ratios because twice the formed surfaces are

required for formwork. Other surfaces, such as the top of the wall and top surfaces of

slabs are not formed. These unformed areas require a “wet” finish, which is applied

by cement masons who strike off and smooth the surface before the concrete takes

final set. Between the elements with high and low ratios are supported slabs and soffit

(bottom) beams, which must be supported by falsework.


TABLE 9.14 Formwork Ratios for Different Elements of Concrete Structures

Element L (ft) Width (ft) HT (ft) Forms (SF) Concrete (CY) Form Ratio

Bridge footings 18 18 6 432 72 6.00

Slab-on grade (SOG),

with CJs 1a100 40 1 280 148 1.89

Small walls 100 20 1 4100 74 55.35

Large walls 100 20 2.5 4100 185 22.14

Bridge piers 6 6 40 960 53 18.00

Small columns 20 5 5 250 19 13.50

Elevated slabs 100 20 1 2240 74 30.24

Misc. curbs & pads 20 20 0.5 40 7 5.40

Totals 12402 632 19.60

aCJs-Construction Joints.

Historical Cost Data Historical cost data are the actual costs or production rates

achieved by a construction company self-performing similar work from project to

project. Costs are recorded as each project is completed, and the final data is used

for determining the project’s financial success and for estimating similar projects in

the future. Successful companies prefer to use their own historical costs over data

published in cost manuals because they represent what the company has achieved on

an actual project. However, published cost and production rates based on documented

equipment and labor rates can be obtained from various sources. The data is based on

an accumulation of records from companies throughout the industry doing the same

type of work. The resulting rates may represent what is expected on an industry-wide

basis, but do not necessarily represent what has been achieved by the estimator’s own

company. Therefore, it is usually preferable for a company to rely on its own historical

cost data when it is available.

Historical costs should take into account safety ordinances, the capabilities and

availability of a company’s supervisory personnel and craft workforce, the area’s

available workforce, and how much similar work the company has previously

performed. Historical costs also include the cost of daily shutdown periods such as

water breaks, 10-min breaks, and bathroom breaks. The production rates generated by

this data take into account all the above, which, when applied to accurate work quan-

tity data, results in a good depiction of actual overall labor costs. These do not include

interruptions by the owner or subcontractors or delays resulting from adverse weather,

labor strikes, or other abnormal interruptions to the work. Published cost manuals that

attempt to take into account all the above using multiplying factors are not company

specific and, therefore, are not as accurate.

Costs Associated with Reinforced Concrete Construction Reinforced

concrete construction requires performance of many different items of work resulting

in several categories of costs. The major costs are permanent materials, formwork

erection and fabrication, small tools and services or supplies (STS), furnishing and

placing reinforcing steel, and placing and finishing of the concrete. Of these, the

ones directly affected by the form ratio are the activities related to the formed contact


area. These specific activities are the labor for erecting and removing formwork, the

placement of the concrete and the dry finishes required to the concrete surfaces.

Formwork Labor Costs Formwork is the largest contributor to concrete laborcosts. As the formwork ratio increases, the costs per cubic yard associated with

formwork also increases. For example, if the cost of formwork per cubic yard ofa structural element is $50 and the form ratio for that element is 15, one wouldexpect the cost of an element with a form ratio of 25 to be higher. Table 9.15 and the

graphic version shown in Figure 9.16 indicate how formwork labor costs increasewith the higher ratios. When such graphs are used in analyzing costs for previouslyperformed work, it is helpful to develop a linear trend line so that the approximate

concrete labor costs can be easily interpolated from a graph similar to Figure 9.16.

TABLE 9.15 Formwork Labor Costs per CY for Different Elements of Concrete Structuresa

Element Type

Form

Ratio

(SF/CY)

Average

Labor Factor

(man-hours/SF)

Cost per SF

($40/man-hour) Cost per CY

Bridge footings 6.00 0.16 $ 6.40 $ 38.40

Slab-on-grade, with CJs 1.89 0.22 $ 8.80 $ 16.63

Small walls 55.35 0.14 $ 5.60 $ 309.96

Large walls 22.14 0.10 $ 4.00 $ 88.56

Bridge piers 18.00 0.08 $ 3.20 $ 57.60

Small columns 13.50 0.15 $ 6.00 $ 81.00

Elevated slabs 30.24 0.25 $ 10.00 $ 302.40

Misc curbs & pads 5.40 0.30 $ 12.00 $ 64.80

Totals (avg) 19.60 $ 7.00 $ 119.92

aCosts are for labor only; materials and equipment are not included.

FIGURE 9.16 Graphical representation of formwork costs per CY.


In addition to the erect and strip labor costs, formwork fabrication labor and mate-

rial costs need to be considered for every square foot of formwork area to be fabri-

cated. As form ratios increase, formwork typically becomes more complicated; this

increases the necessity for forms, which reduces the reuse factor. As the formwork

ratio increases, the form reuse factor decreases. The reuse factor has to do with how

many times one form can be used before it is thrown away or modified significantly

(erect and strip quantity divided by form fabrication quantity), or simply the number

of times a form is reused. If the form reuse factor is low, the project must produce

more forms. For example, if the formwork material costs $5.00 per SF and they can

be fabricated at a labor rate of 0.08 man-hour/SF (with labor @ $40.00/man-hour),

the project would incur an additional cost of $8.20 per SF on form fabrication.

Another significant phenomenon is the trend that occurs in placing and removing

forms by the square foot as the ratio changes. For small and detailed structures, when

the ratio increases, the man-hours required per square foot decreases. In some sample

past cost analyses of a form erect and strip operation, a form ratio of 16 yielded 0.195

man-hours per square foot. As the form ratio reached 34, the demand on erecting and

removing the forms reduced to 0.148man-hours per square foot. This is caused by the

amount of area produced by the forms being different for the same amount of effort.

Concrete Placement As mentioned earlier, low form ratios are typically associ-

ated with slab-on-grades and thick elements, particularly those that are formed on one

side only.When concrete is placed in these elements, more concrete can be placed per

hour for the same size placing crew. The more concrete placed per hour, the less the

labor hours required per cubic yard. As a comparison, high form ratios are associated

with thin walls and elevated slabs and beam/girder elements (falsework supported).

Thin walls have to be poured slowly, therefore, for every vertical foot of concrete,

fewer cubic yards are placed. This can be compared to wider walls, which require

much more volume for every vertical foot of placement. If a wall thickness dou-

bles, the ratio and the cost per cubic yard are reduced to half. Elevated slabs and

beam/girder elements produce slower concrete placements because they are typically

thinner than on-grade element, and they sometimes have to be placed in a partic-

ular sequence to satisfy falsework design requirements. All forms are above-grade

designed with specific pour rates, which must be followed to avoid form failures.

Concrete conveyance costs go hand in hand with placement costs. An example

situation in which the concrete is conveyed to the forms by a concrete pump/boom

system costing $180/h and is placed at a rate of 72 CY/h would require adding $2.50

per CY (180/72) to the placing costs. The longer a pour takes, combined with how

many cubic yards are actually being placed per hour, greatly affects the placement

cost. Consequently, if the pump is on site longer, the placing crew is also there longer.

The setup and cleanupwould be the same at the beginning and end of the pour, regard-

less of the time required to place the concrete.

Concrete Finishes Finishes in concrete work have two common classifications:

wet and dry. Wet finishes are applied shortly after the concrete is placed but prior to

the initial concrete set. Dry finishes are required on formed surfaces after the forms

are removed in order to achieve the requirements of the specifications for smooth-

ness and other aesthetics. Wet finish costs do not fluctuate with form ratios as dry


finishes do. Dry finishes have a direct relationship to the formed surfaces. Formed

surfaces will require some type of dry finish to meet the specification. Almost all

specifications require either a point and patch or sack finish. Surfaces that will not

be exposed to view when the structure is put into service generally require a much

less stringent dry finish requirement. These types of finishes would also apply to sur-

faces that are backfilled and buried underground. The labor costs for these finishes

range from 0.008 to 0.015 man-hours per square foot, depending on the specification

requirements. On the other hand, concrete surfaces exposed to view, such as the out-

side surfaces of structures and public area exterior surfaces, will produce labor costs

from 0.02 to 0.06 man-hours per square foot. Therefore, when the form ratio fluctu-

ates from a 5 to a 10, the finish costs will range from $1.00–$10.00 per CY ($40.00/

labor-hour).

Architectural finishes that are formed typically require a more expensive form and

dry finish. There are also additional STS costs for the form liner material to produce

the different patterns in the concrete. These form liners are attached to the face of the

formwork in order to achieve the architectural design surface. Various patterns avail-

able on the market cost between $5.00 and $40.00 per SF. The inexpensive material

is hard PVC and the higher quality, more expensive material is made from an elas-

tomeric and can resist wear from multiple concrete placements. Form liner costs vary

directly with the surface area and are related to the form ratio.

Small Tool, Services, and Supplies (STS) In addition to the labor costs for rein-

forced concrete estimates, small tools and services required to support formwork

materials must be considered for purchase. This category includes formwork place-

ment tools and incidentals and concrete placement and finishing tools. Formwork STS

includes plywood, dimensional lumber, form ties, and hand power tools. Concrete

placement STS includes vibrators, shovels, curing hoses, fittings and blankets, rub-

ber gloves and boots, and construction joint preparation equipment such as sandblast

pots, nozzles, hose, respirators, and goggles. Finishing STS include trowels, floats,

cement and sand for dry patch, and personal protective equipment.

Formwork ratios fluctuate up and down, which affects the items associated with

SFCA. Of the above-mentioned items, certain ones are directly related to the square

foot of contact area (SFCA). These are formwork fabrication, placing, and removing

formwork, dry finish, and wet finish. Typical costs for these items are:

Form fabrication: $1.50–$30.00∕SF of fabricated forms

Form erect and strip: $0.30–$0.40∕SF for all forms that are placed and removed,

regardless of reuse

Concrete placement: $1.70–$2.00∕CY of concrete placed

Dry finish concrete: $0.10–$0.15∕SF of concrete, which requires some type of dry

finish

Higher form ratios produce higher STS costs per cubic yard and labor hour.

For example, a ratio of 5.0 would result in $1.50–$2.00∕SFCA of formwork STS

($0.30∕SF × 5.0 or $0.40∕SF × 5.0). On the other hand, a form ratio of 20.00 would

produce $6.00–$8.00∕SF of formwork erect and strip, respectively ($0.30∕SF × 20.0


FIGURE 9.17 STS cost as related to form ratio.

or $0.40∕SF × 20.0). Figure 9.17 shows the relationship between STS costs andconcrete labor man-hours. The trend line indicates the increase in STS costs with anincrease to the form ratio. The data points represent past project costs.

Reinforcing Steel Reinforcing steel can vary by structure type and project.However, when concrete ranges from 8 to 60 in in thickness, the reinforcing steelratio (pounds of steel per cubic yard of concrete) can vary in a similar fashion to theform ratios. For instance, a 12-in wall will typically have reinforcing steel at bothsurfaces (1 1

2′′ to 2′′ from the surface). In comparison, a 5′0′′-thick wall or column

would also have two planes of reinforcing steel, except this larger element wouldhave at least 4′0′′ of unreinforced concrete between the two layers of reinforcingsteel. The 12-in wall would have a much greater weight of reinforcing steel per cubicyard than the larger element. If it is assumed that these two elements have a formratio of 5–20, then it would be expected that the reinforcing weight per cubic yardwould increase with the ratio. As an example, a 6-ft2 column that is 20 ft tall andreinforced with 5-lb bars 2 in from the face, all around, and 12 in on center in eachdirection would produce 36 lb of rebar for every cubic yard of concrete. However, a12-in-thick wall, 6 ft long and 20 ft tall with the same reinforcing layout produces108 lb of rebar for every cubic yard of concrete. Structures that are more heavilyreinforced would achieve higher ratios proportionately.

When a concrete quantity takeoff is complete, an estimator can “plug” a dollaramount for the reinforcing subcontractor. To do this, he often relies on the volumeof concrete and a review of the size and spacing of the reinforcing steel. The sizeand spacing is going to indicate to an estimator approximately how many pounds ofreinforcing steel may be needed for every cubic yard of concrete, similar to the aboveexample. Once the weight is estimated, the cost (plug) can be obtained by review-ing historical project data that shows costs ranging from $0.75 to $1.25 per poundof placement ($1.10 to $1.75 if the reinforcing steel is epoxy coated). Since rein-forcing is traditionally performed by a subcontractor specializing in furnishing and


TABLE 9.16 Portions of Concrete Costs Related to the Form Ratioa

Element Type

Form

Ratio

(SF/CY)

Erect &

Strip

Labor STS

Place &

Finish

Labor

Total Cost

per CY

Bridge footings 6.00 $120.00 $11.50 $24.00 $155.50

Slab-on-grade 1.89 $140.00 $5.34 $7.56 $152.90

Small walls 55.35 $320.00 $85.53 $221.40 $626.93

Large walls 22.14 $180.00 $35.71 $88.56 $304.27

Bridge piers 18.00 $160.00 $29.50 $72.00 $261.50

Small columns 13.50 $360.00 $22.75 $54.00 $436.75

Elevated slabs 30.24 $400.00 $59.86 $120.96 $580.82

Misc curbs & pads 5.40 $375.00 $10.60 $21.60 $407.20

Totals (avg) 19.60 $365.73

aMargin and indirect costs are not included in the above estimates.

installing reinforcing steel, these costs are left out of the overall labor costs normal

to the concrete labor estimate.

Overall Costs The costs mentioned so far are estimated as individual components.

Once they are analyzed individually, they can be combined into total costs. Table 9.16

shows some sample costs for the total concrete estimate. The total costs are an

accumulation of the costs previously mentioned that fluctuate with form ratios. For

instance, the purchase of the actual concrete has been left out since the volume of

concrete is not affected by the form ratio by itself. Margins (company profits) and

indirect costs were also left out of these costs. In other words, the costs shown are an

estimate of only the direct labor costs that are affected by form ratios.

Types of Structures Structure types vary between the different uses of the struc-

tures. The geometry, size, and shape of each structure determine the form ratio. The

following are categories of concrete structures and their elements:

Highway Bridges: footings, abutments, piers, and superstructure

Water/Sewage Facilities: slabs-on-grade, walls, and elevated slabs

Commercial and Industrial Buildings: slabs-on-grade, columns, walls, and ele-

vated slabs

Bridges in California, as in many other western states, are typically the concrete

box-girder type. The substructure (footings and columns/piers) and the superstructure

(spans from pier to pier) are all cast-in-place reinforced concrete with posttensioning

strand. The substructure form ratio is below 12. The superstructure form ration is

above 20. Water treatment structures and commercial and industrial buildings have

form ratios greater than 18. The foundations and walls of the latter are less thick and

exhibit elements that do not have to support high dead loads; therefore, they consist

of smaller elements. Refer back to Table 9.16 for typical form ratios as they apply to

these elements.


When estimating different types of structures, the estimator should categorize thetypes of elements within the project type and keep these cost estimates separate. Thiswould eliminate the chances that costs for the low ratio structures are not mixed withcost of the higher ratio structures. Once the elements are categorized, the quantitytakeoffs should remain separate and only combined when intending to represent theoverall project.

Estimating with Formwork Ratios

The Work Quantity Takeoff and Labor Work-Hours The construction specifi-cations will assist the estimator in determining how to categorize the takeoff in thecost-estimating process. A quantity takeoff is performed for all the individual workitems that involve labor and materials. It is recommended that the structure typesand the elements of the structures be categorized prior to the actual takeoff. Once itis determined which categories contain which structures and their specific elements,the estimator can begin quantifying the square feet of form contact area, the cubicyards of concrete, the dry and wet finishes, and the like. Spreadsheets are often usedto break the elements down into components within an element. The spreadsheet(whether done by hand or with a computer) should show subtotals for each element,and for each structure type. It is very common for a company to require two separatequantity takeoffs, performed by two different takeoff engineers, when projects arebid by lump sum. This assures accurate quantities and reduces the risk exposure tomonetary loss.

When the structure takeoff is complete, the form ratio for each of the work activ-ities can be calculated. Then the estimator must decide on the man-hour productionrate for each of the work activities that must be performed for each structural element.Table 9.17 shows an example of tabulated slab-on-grade historical costs data sheet.

If an estimator was pricing slab-on-grade formwork, data would be tabulated asshown in Table 9.17. Three past projects are listed next to an actual work quan-tity for the project being estimated. Historical production rates of 0.185, 0.124, and0.234 man-hours per square foot are shown in the table as an example. The estimatormust then decide what production rate to apply to the present project slab-on-gradeactivity. The three past projects can be used to determine the appropriate produc-tion rate. One way to do this would be to average the three past project productionrates. However, this method would not consider the differing work quantities of thethree projects. For instance, if the first project required 2000 ft2 of forms, the second

TABLE 9.17 Example of Completed Project Quantity and Labor Hour Factors

This Estimate Past Project 1 Past Project 2 Past Project 3

Description QTY UNIT

Man-

hour/

SF

TOTAL

Man-

hours QTY

Man-

hour/

SF QTY

Man-

hours/

SF QTY

Man-

hours/

SF

Erect and strip

Slab-

on-grade

forms

3200 SF 0.400 800 2000 0.185 5500 0.124 350 0.234


5500 ft2, and the third 350 ft2, an averagewould not reflect the effects of their differing

quantities.

What is the effect of differing quantities? Larger work quantities such as those

for the project 2 with 5500 ft2 typically produce a better production rate because the

operation generally continues to improve with time. On the other hand, project 3 with

350 ft2 did not experience enough repetition to refine the process and produce better

unit rates. In this case, it is not a surprise that the unit rate of 0.234 exceeds the others.

The 2000-ft2 project 1 falls somewhere between the other two.

If the project being estimated has a slab-on-grade quantity of 3200 ft2, and the

estimator has no reason to believe that the project elements are much different from

the projects in the past cost summary, then the estimator would look at the two larger

quantities (5000 ft2 and 2000 ft2) and choose a production rate somewhere in between.

In this case, it would be appropriate to interpolate the unit rate between the other two,

resulting in a rate of 0.163 man-hours per square foot.

When this is done, the results for each structure element can be totaled to rep-

resent the entire project. These totals will include the total cubic yards of concrete

and the total square feet of contact form area resulting in the total labor hours per

cubic yard of concrete for a given form ratio. The corresponding information for the

present project can then be plotted for a comparison between the historical rates and

anticipated performance. If the plots for the present project fall above or below the

“trend line,” then it is evident that the estimate is either high or low when compared

to historical data.

When the activities have all been compared to past costs individually, and their

production rates have been determined, another analysis should be done. Since we

know the form ratio for the current estimate and the total concrete quantities, we

should be able to line them up with historical costs of other projects and their totals.

By doing this, similar projects are being compared. It would not make sense to com-

pare costs between a project with a form ratio of 12 to a project with a form ratio

of 22. This would defeat the purpose of using these ratios and comparing job totals.

The form ratios of the project being compared should be within 10–15% higher or

lower than the trend line of similar projects. Greater variations are cause for concern

with regard to the present estimate. It should be noted that the above analysis per-

tains only to cost of labor and associated STS. A similar analysis can be performed

for other concrete costs.

Tracking Project Costs Everything that has been discussed so far would be irrele-

vant if the project staff did not do an accurate job of tracking costs at the project level.

A system should be in place to record quantities completed, labor hours spent, and

total activity costs including equipment and materials. A system like this is almost

always used so that companies can document the actual costs as compared to the

estimated budget.

In addition, it is very important that the project staff accurately code the cost of

individual activities. Project personnel are sometimes tempted to make an activity

seem more cost effective than it is or claim quantities for work not actually per-

formed, thus producing inaccurate results. The results, if used on future estimates,

would generate an unrealistic estimate and eventually jeopardize the financial health


of the company. The project reporting system needs tomatch the estimating system by

category. These categories are similar to the estimating activities mentioned earlier.

9.3 CONCLUSION

Many items were covered in this chapter between formwork design and formwork

costs. Since formwork can make up to 75% of the concrete costs, it cannot be empha-

sized enough how important the design of the forms are to these total concrete costs.

Concrete estimates (including formwork) are performed in similar manners from

company to company. The student of this subject should leave academia with a solid

understanding of the components between design and costs. Then, when putting the

studies to practice, learn the actual methods used by the company in which the student

is employed.

CHAPTER 10

FALSEWORK DESIGN

Chapter 9 set the stage for basic formwork calculations and the three standard design

checks: bending, shear, and deflection. As we learned with formwork, concrete struc-

tures leave a permanent impression; and in many cases the final product is viewed by

the public, Therefore, in its temporary nature, concrete not only has to comply with

bending and shear issues, but concrete cannot deflect or undulatemore than the human

eye can see. Falsework is similar, but it is easier to see these undulations in vertical

construction (walls and columns) than it is to see in horizontal constructions. With

that said, let’s bring to the table the concepts of risk and catastrophe. It is one thing to

build a product that is not visually acceptable or pleasing to the eye. However, when

constructing elevated, horizontal concrete segments, we must contend with gravity,

and a whole other element comes into play.

10.1 FALSEWORK RISKS

Falsework supports elevated horizontal concrete elements such as building slabs,

bridges, and beams. For years, owners and engineers have recognized falsework as

being a temporary process to construction that needs to be engineered by a licensed,

professional engineer. For years, vertical construction was not in this category, but

in recent years, even vertical formwork has fallen under the scrutiny of the owner

and his engineer. No owner wants any sort of incident associated with the project, no

matter how large or small. Chapter 11 will get into design issues as they pertain to

pipe falsework; and it can be assumed that these same issues would also be of concern

for smaller horizontal and vertical concrete works. Figure 10.1, for example, shows

falsework over a river using pipe post falsework. The risks of this type of work are

illustrated from fall risks, equipment risks, and environmental risks at a minimum.

229

230 FALSEWORK DESIGN

FIGURE 10.1 Pipe post falsework.

The intent of this book is not to prepare the student for designing and building

formwork and falsework to construct the new San Francisco/Oakland Bay Bridge,

the Tappan Zee Bridge in New York, or a 100-story downtown office building. The

intent here is to educate the student for designing and building safe systems that are

economically efficient and within the project’s budget. The larger projects will be

experienced eventually when the student builds confidence and is surrounded by a

team of experts in their firld.

10.1.1 Falsework Accidents

It seems that a year never goes by in the United States without a report of a false-

work accident. These accidents are not just the result of design problems but are

often caused by human error or bad decisions made by field supervisors from super-

intendents to foremen and below. Even though methods and procedures are just as

important, if not more important than design factors in temporary structures, the focus

of this book will remain on design. In any case, whatever their cause, accidents are

always unfortunate; and the only good that comes from them is the inherent lesson to

be learned.

CASE STUDY OF A FALSEWORK ACCIDENT

Project Overview

Soon after the turn of the 21st century, a highway project was constructed

in northern California by a joint venture company for the State of California

(owner agency). The project involved road widening, overhead bridge ramps,

10.1 FALSEWORK RISKS 231

and accompanying drainage on Highway 149 in Butte County. Figure 10.CS1

shows typical frames (bents) and stringers (beams) in place on this project for

the typical pipe falsework system.

FIGURE 10.CS1 Typical falsework for this project.

The completed project would eliminate the need for surface intersections

with stop signs or signals. On July 31, 2007, an overpass that was under

construction collapsed.

Fortunately, no one was killed as a direct result of the collapse. However,

one fatality resulted from traffic buildup, and one construction worker and one

person driving underneath the collapse were injured. This relatively fortunate

outcome was not the result of good decision making or good planning, but was

pure luck. The contractor and the State of California dodged a deadly and costly

bullet.

Forensic studies were performed, as one would expect, by Cal OSHA.

However, the cause of the accident was not ultimately decided by a judge.

No one served any time in prison. The injured persons settled with the contrac-

tor and insurance companies; and by 2011, one would have thought nothing

had ever happened. So what actually caused the collapse of the falsework?

The Cause

A falsework bent consisting of pipes and beams collapsed onto a passing truck,

nearly crushing the driver and causing a construction worker to fall, when a

(continued)


CASE STUDY OF A FALSEWORK ACCIDENT (Continued)

guying cable was removed. The frames and beams were placed during a previ-

ous night shift with the highway closed in both directions with traffic detoured.

During this shift, the frames were braced with wire rope guys to temporary

deadmen (counterweights). Before the wire rope guys could be transferred to

the adjacent bent for a more permanent attachment, the shift had ended and the

highway was reopened to traffic. The following morning shift began, and the

crew and owner agency realized the guying was not complete. Instead of set-

ting up another road closure the following evening and correcting the situation

without live traffic, the decision was made by the contractor and owner agency

to relocate the guys that morning.

When the guy was released, so much tension had built up that the frame

moved enough to generate momentum that toppled the frame and stringers it

supported. The falsework bent failure was caused by human error. Had the plan

been followed, the accident would probably have been avoided. However, had

a better bracing system been designed and installed initially, this could have

prevented the accident.

Corrective Measures

The project was delayed for several weeks. Once the investigation allowed the

contractor to continue, a new bracing system was put into place. Figure 10.CS2

shows the new pipe bracing system in place.

FIGURE 10.CS2 Push–pull braces added to falsework.


This bracing was added to the new falsework bents that were erected after

the accident. They went to a push–pull system to take the wire rope tension out

of the equation and to allow bracing on only one side of the bent since traffic

was on the other side.

The parties involved spent several years, after the completion of the project,

in settlement negotiations. The settlements were never made public.

The next section will discuss the measures that are taken by falsework contractors

and owner/engineers to assure that accidents due to either design or to human error

or incorrect methods are avoided.

10.1.2 Falsework Review Process

Two types of falsework review procedures are common in the construction indus-

try. These review procedures are engineering checks and balances and they are

(1) in-house reviews and (2) owner/engineer reviews.

In-House Reviews In-house reviews occur within the company performing the

falsework erection and removal. This part of the process can, and frequently does,

involve consulting engineers when the contractor does not have an engineer on his

payroll or its engineering staff is too busy. Regardless of the engineering component,

this group represents the contractor.

Some contractors develop written procedures that are followed during the design

process, prior to submitting it to the owner/engineer. Thesewritten procedures contain

some of the following information:

Type of falsework risk

Eligibility of persons involved

Redundancy

Risk avoidance

The contractor must categorize the risk somehow. An example would be to rate

the falsework type as low, medium, or high risk. Once the levels have been decided,

each risk level would represent a specific type of falsework. In other words, the

lower risk falsework would be used for the most straightforward, easier systems;

higher risk falsework would be used for projects spanning sensitive waters, busy

highways, or in very tall systems. Table 10.1 has been prepared as one of many

options for a review policy matrix. The dimensions used are recommendations or

estimates. Each company should consult its legal department before preparing a

similar policy.

Each company should develop a policy, communicate that policy to the employees,

andmanage the policy. Obviously, some terms would still have to be carefully defined

in the document. For instance, the word “minimal” could mean different things to

different people. The policies cannot have vague wording or contradictions.


TABLE 10.1 Example of Falsework Risk Categories

Risk Category

1. Designer and

2. Check Engineer Falsework Type

Low 1. In-house engineer, consultant

engineer

2. In-house engineer, consulting

engineer (not the same)

Elevation less than or equal to 20 ft in

height, beam spans less than or equal

to 30 ft in length. No risk to public; no

risk to environment.

Medium 1. Consultant engineer

2. In-house engineer, consulting

engineer (not the same)

Elevation less than or equal to 30 ft in

height, beam spans less than or equal

to 40 ft in length, minimal risk to

public, some minor roadways,

minimal risk to environment.

High 1. Consultant engineer

2. Additional consultant engineer

Elevation greater than 30 ft in height,

beam spans less than 40 ft in length,

high probability there is risk to public,

such as highways, railroads, and

private property, high probability

there is risk to environment, both

which would mean substantial

financial company losses.

Persons who are eligible to design and check the design should be clearly speci-fied as well. In no case should these two responsibilities be performed by the sameperson. Also, in no case should these two responsibilities be performed in-house.In some cases (high risk), the company may not want any in-house involvement.The risk, in this case, would be 100% subcontracted to a consulting firm. Each com-pany should evaluate its own risk and write a specific policy that addresses the mostcritical management concerns.

Owner/Engineer Reviews On the other side, that of the owner/engineer, simi-lar concerns are shared for possibly different reasons. No owner/engineer wants anyproblem of any kind to occur during the construction process. Since falsework comeswith additional risks from normal construction activities, it is scrutinized to a higherdegree. Falsework approvals are, in fact, so important to CalTrans that the CalTransStandard Specifications manual contains a separate specification section for false-work. Their specifications cover anything that has or could possibly go wrong on ahighway project.

Owner/engineer reviews can contain a great deal of requirements. The followinglist is just some items that might be required in a falsework submittal:

Material compliances and specifications

Letters of certification

Welding certifications for previously welded splices

Calculations for all elements

Shop drawings, separate submittals for each bridge and each frame

Erection and removal procedures, with sequencing


Material grades and species

Anticipated settlements

Required experience of designer when high risk (minimum 3 years)

10.1.3 Falsework Design Criteria

The basic design checks that apply to both formwork and falsework are:

Bending stress for plywood and dimensional lumber, l = 10.95√FbS∕w

Shear stress for plywood, L = Fv(Ib∕Q)∕0.6wShear stress for dimensional lumber, L = [F′

v(b)(d)∕0.9w] + 2d∕12Deflection for plywood and dimensional lumber, l = 1.69 3 3

√EI∕w, l is clear span

in inches, and L is clear span in feet.

If a form designer is using plywood and dimensional lumber for all falseworkcomponents, these formulas will be instrumental. When other materials are used, aswill be the case later in this chapter, other formulas and charts will be necessary.

Falsework loads, even though originating from material weight and gravity, areactually fairly light compared to hydrostatic pressure. In the last chapter, the loadsfrom hydrostatic pressure were in the range of 800–1200 psf. Many falseworkloads, except for large public works and transportation projects, are in the range of200–500 psf. As mentioned in Chapter 8, falsework loads include dead loads and liveloads and self-weight of stringers. However, even with these additional forces, mostfalsework loads are still lighter than formwork loads. As contractors, we need to becontinuously reminded of the risk in construction of elevated horizontal concreteworks and always perform the checks and balances that go along with this type ofwork that were detailed earlier in this chapter.

This chapter will show three examples of different types of falsework in construc-tion. The first two will be standard wood falsework supporting elevated reinforcedconcrete slabs. One will use the conventional method for calculating loads, and theother will use the Appendix 7 formwork charts used in the previous chapter. The firstexample will also incorporate the load calculations learned in Chapter 8 so the stu-dent can continue to master this process. Example 10.3 will utilize aluminum beamsand 15-k shoring frames.

As mentioned in Chapter 8, elevated slab loads are a combination of the concreteand material dead load (DL) and the live load (LL) from workers, equipment, tools,and the weight of the falsework beams. The wood and plywood weight are includedwith the DL. The formula given was Pmax = DL (concrete, rebar and forms) + LL(workers, equipment, tools and steel beams).

DL = Slab thickness (feet) X 160 pcf (160 = concrete unit weight including

rebar and forms)

LL = Depends on the type of construction (ranges from 25 to 75 psf)

Light duty = 25 psf (painters and cleaning)

Medium duty = 50 psf (carpentry)

Heavy duty = 75 psf (concrete, masonry, demolition, and steel)


FIGURE 10.2 Water tank falsework using aluminum system.

10.1.4 Load Paths for Falsework Design

The following paths and component terms will be used in this chapter, depending on

the type of construction. Following the example types, the load paths described will

be flat slab with wood posts; flat slab with dropped beams using aluminum joists,

beams, and frames; and bridge falsework using wood and steel beams.

1. Elevated slab support with all wood components—plywood, joist, cap beam,

post, ground, or lower slab

2. Elevated slab support with aluminum components—plywood, aluminum joist,

aluminum cap beam, aluminum 15-k frames, ground or lower slab (Figure 10.2)

3. Bridge falsework with a combination of wood and steel beams—plywood,

wood joists, steel stringers, steel cap beams, posts, lower cap beams, sandjacks/

wedges/corbels, pads, or ground slab

Example 10.1 Flat Slab FalseworkDesign falsework to support a 25′′ slab with heavy-duty live load. In order to save

time, deflection will be ignored in these first two examples. Use Douglas fir–Larch,

4 × 4 joists, and 4 × 8 cap beams.

Step 1: Determine the loading from the DL + LL (see Figure 10.3).

DL = 25′′∕12′′∕ft X 160 pcf = 333 psf

LL workers∕equipment = 75 psf

Total = 408 psf (use 410 psf)


FIGURE 10.3 Load diagram of example 10.1.

Step 2: Select the plywood and determine the joist spacing.

Since the plywood will be placed by hand and the loading is relatively light com-

pared to wall forming, use5

8′′ BB-OES in the strong direction. The plywood has an

allowable bending stress value of 1600 psi and allowable rolling shear value of 60 psi.

W = 410 psf × 1 ft = 410 plf

Bending

l = 10.95√(FbS∕w)

where l = clear span (in)

Fb = allowable bending stress of material (1600 psi)

S = section modulus (0.339 in3)

W = uniform load on span (410 plf)

10.95 = constant

l = 10.95√(1600)(0.339)∕410) = 12.59 in

Shear-Plywood

L =Fv(Ib∕Q)0.6 w

where L = clear span (ft) divide by 12 to compare in inches

Fv = allowable shear stress of material (60 psi)

Ib∕Q = rolling shear constant (5.824 in2)

W = uniform load (410 plf)

0.6 = constant

L = (60)(5.824)0.6(410)

× 12′′∕ft = 17.04 in

Deflection is ignored for this example because it only controls long spans with

light uniform loads.

l = 1.69 3√(EI∕w)

where l = clear span (in)



E = modulus of elasticity

1.69 = constant


Ignoring deflection, the joist spacing can’t exceed 12.6 in of clear span, which

is governed by bending. The on-center spacing of the joists will be greater, and

the amount depends on the width of the joist. If a Douglas fir–Larch (DFL) 4 ×4 (b = 3.5′′) joist is used, then the on-center spacing would be 12.6 + 3.5′′ = 16.1′′,say 16′′ OC. Figure 10.4 shows this spacing between joists.

FIGURE 10.4 Joist spacing for example 10.1.

Step 3: Determine the joist size and cap beam spacing.

In step 2, the joist size was selected to be a DFL 4 × 4. This joist size will be

used to determine the cap beam spacing. First the uniform load on one joist must be

recalculated.

W = 410 psf × (16′′∕12′′∕ft) = 546 plf

along each joist. With this uniform load and the properties of the DFL 4 × 4, deter-

mine the cap beam spacing. Perform the two checks, ignoring deflection.

Bending

l = 10.95√(1000)(7.15)∕546 = 39.63 in


As a reminder, the dimensional lumber calculation for shear is different than plywood:

L =(F′vbd

0.9w

)+(2d12

)

where L = clear span (ft), divide by 12 to compare in inches

F′v = twice the allowable shear stress (2 × 95 psi = 190 psi)

b = base dimension of wood section (3.5 in)

d = depth dimension of wood section (3.5 in)

w = uniform load (546 plf)

0.9 = constant

L = 190 × 3.5 × 3.5

0.9(546)+ 2 × 3.5

12= 5.32 ft

l = 5.32 × 12′′∕ft = 63.8 in

Deflection is ignored for this example because it only controls long spans with

light uniform loads.

Ignoring deflection, the cap beam spacing can’t exceed 39.6 in of spacing, which

is governed by bending. Space the cap beams at 39′′ OC because 39′′ is 3.25 ft, and


this will normally work out better for the overall layout of the falsework. Figure 10.5

shows the cap beam spacing.

FIGURE 10.5 Cap beam spacing for example 10.1.

Step 4: Determine cap beam size and post spacing.

In step 3, the cap beam size was selected to be a DFL 4 × 8. This cap beam size

will be used to determine the post spacing. Once again, the uniform load on one cap

beam must be recalculated.

W = 410 psf × (39′′∕12′′∕ft) = 1333 plf

along each cap beam. With this uniform load and the properties of the DFL 4 ×8, determine the post spacing. Perform the two checks, again ignoring deflection.

Be sure not to double the cap beam properties as was done in formwork using double

walers. Cap beams are single beams, so the properties are left single.

Bending

l = 10.95√(1000)(30.66)∕1,333 = 52.51 in

Shear

L = 190 × 3.5 × 7.25

0.9(1333)+ 2 × 7.25

12= 5.23 ft

l = 5.23 × 12′′∕ft = 62.8 in

Deflection is ignored once again.

Ignoring deflection, the post spacing can’t exceed 52.5 in of spacing, which is

governed by bending. Space the cap beams at 52′′ OC. Figure 10.6 shows the cap

beam spacing.

FIGURE 10.6 Post spacing for Example 10.1.


Step 5: Determine post load and sizing.

The post load can be determined by distributing the Pmax load over the tributary

area of one post.

Ppost = Pmax × (cap beam spacing × post spacing)

P = 410 psf × (39′′∕12′′∕ft) × (52′′∕12′′∕ft) = 5,774 lb

The tributary area per post is 14.08 (39′′ × 52′′∕144) ft2. A post must be designed

that can support 5774 lb. For this example, it is assumed that the post height (from

ground to bottom of cap beam) is 13 ft. This represents Le = effective post length.

With this information, a 4 × 4 DFL post can be checked. If this post does not work,

a larger size should be checked or the post spacing can be decreased to lower the load

per post. Typically, Le can’t be changed because the elevated slab must remain at the

same elevation per the contract plans. However, Le can be reduced if lateral bracing

is added. For this example, it is assumed that lateral bracing will not be used, and the

Le will remain at 13 ft.

To check the post size, perform the following check:

PA

>0.3E

(Le∕d)2

where P∕A = normal axial stress on the post

E = modulus of elasticity of DFL

Le = effective length of the post in inches

d = least dimension in a cross section of the post

Normal stress on the post would be

PA

= 5774 lb

(3.5)(3.5)= 471 psi

This is the normal stress at the end of a 4 × 4. The allowable compression stress

parallel to the grain for DFL is 1600 psi, which is well below the crushing limits.

Buckling is the main issue when a 4 × 4 is required to support a load as a 13-ft-long

post. To do this, use the formula above to determine the allowable buckling stress and

ultimately the allowable load the post can support:

PA

>0.3(1,500,000)

(13′ × 12′′∕ft)∕(3.5′′)2= 227psi

The allowable buckling stress for this condition ismuch less than the normal stress;

therefore, the buckling stress dictates the design of the post. To determine the allow-

able post load, plug the buckling stress back into fbs = P∕A, solve for P, and compare

this to 5774 lb.

Pall = 227 psi × (3.5 × 3.5) = 2781 lb < 5,774 lb (does not work)

Since the allowable load due to buckling is not equal to or greater than the actual

load of 5774 lb, a larger post must be considered. Before going to a 6× (which is very


expensive), the designer should try a 4 × 6, 4 × 8, and so forth. The largest 4× that

should be considered is the size just below the cross-section area of a 6 × 6 = 36 in2

(which would be the first 6× to be tested). A 4 × 8 is 32 in2 and a 4 × 10 is 40 in2;

therefore, if a 4 × 8 does not work, then the 6 × 6 should be considered. Since the

d dimension of both a 4 × 6 and 4 × 8 are 3.5′′, the allowable buckling stress can be

multiplied by the cross-sectional area to quickly calculate the allowable load on each.

In other words, there is no need to recalculate the allowable buckling stress, as it will

not change as long as the least dimension, d, does not change:

Pall for 4 × 6 = 227 psi × (3.5′′ × 5.5′′) = 4370 lb < 5774 lb (does not work)

Pall for 4 × 8 = 227 psi × (3.5′′ × 7.25′′) = 5760 lb < 5774 lb (does not work,but very close)

By observation, it appears a 4 × 10 would work, but for economy purposes, it was

pointed out that a 6 × 6 should be used if a 4 × 8 did not work. Unfortunately, now,

the buckling stress must be recalculated due to the increased d dimension changing

from 3.5′′(4×) to 5.5′′(6×).

Fbs = 0.3(1,500,000 psi)∕(13 × 12′′∕ft∕5.5)2 = 559 psi

Pall = 559 psi × (5.5 × 5.5) = 16,910 lb > 5774 lb (OK)

Usually when the post goes up to the next minimum dimension, the allowable

capacity doubles or triples. Also, if the designer really wanted to use the 4 × 4’s, he

could go back to the beginning, then add lateral bracing and decrease the Le.Anotherway to make the 4 × 8 work would be to slightly decrease the spacing between either

the posts or cap beams, thus reducing the tributary area per post and eventually the

post load. At this point, a cost comparison should be done between the 6 × 6 and the

labor and material cost for the lateral bracing (lacing). Finally, the savvy student has

also recognized that since we used a 6 × 6, the b dimension used in step 4 should be

increased to 5.5 in, which changes the post load. Technically, this slight change can

be ignored because, by using the 6 × 6, we built in an additional factor of safety =16,910∕5774 = 2.93 FOS.

Bearing Capacity

The crushing stress at the end of the 4 × 4 was discussed, but the crushing between

the 4 × 8 cap beam and the 6 × 6 post has not been checked. This is more critical

because this crushing is considered perpendicular to the grain of the 4 × 8 cap beam,

which is more sensitive than the end of the post (parallel to the grain of the post).

Figure 10.7 illustrates the bearing area in question.

fc⟂ = P∕A(rea)

fc⟂ = 5774 lb∕19.25 in2 = 300 psi

The allowable crushing stress perpendicular to the grain for DFL is 625 psi >

300 psi (OK).


FIGURE 10.7 Bearing area between the post and cap beam.

Final Design:

5/8′′ BB-OES plywood used in the strong direction

DFL 4 × 4 joists spaced at 15′′ OC

DFL 4 × 8 cap beams spaced at 42′′ OC

DFL 6 × 6 × 13 ft long

No lacing for posts, crushing in both directions (OK)

10.1.5 Falsework Design Using Formwork Charts

The next example will utilize the formwork charts in Appendix 7. The charts selectedin this chapter are from theWilliams Form Catalog. These particular charts use fairlyliberal shear values, hence all spacings are governed by either bending or deflection.

Example 10.2 Flat Slab FalseworkUsing the formwork tables in Appendix 7, design falsework to support a 21′′ slabwith medium-duty live load. Deflection will still be ignored.

Design a falsework system using the charts in Appendix 7 and following these

parameters:

DL = 21′′∕12′′∕ft X 160 pcf = 280 psf

LL workers∕equipment = 50 psf

Total = 330 psf (Figure 10.8)

FIGURE 10.8 Slab load diagram.

Step 1: Plywood selection and joist spacing.

For this example, 3∕4′′ BB-OES plywood will be used. Based on this plywood and

the charts in Appendix 7, determine the joist spacing. This appendix is using material

that has the following characteristics:

Max deflection L/360, not to exceed1

16′′

Fv = 72 psi


Fb = 1930 psi

E = 1,500,000 psi

From the previous calculation, w = 330 psf

Based on the plywood being used with the face grain parallel to the span (left

side of chart) and w = 330 psf, the joist spacing should not exceed 17′′ OC. In orderto make the fabrication and erection most productive, the joists will be spaced at

an even 16′′ OC. Like the charts in the previous chapter, the values given are the

center-to-center spaces.

Step 2: Joist selection and cap beam spacing.

Uniform joist load: w = 330 psf × (16′′∕12′′∕ft) = 440 plf

The next chart in Appendix 7 that is used is intended for all single-beam (studs,

joist, and others) members. Also, this chart represents beams that are supported by

three or more supports (M = wl2∕10), not simply supported (M = wl2∕8.).In this example, a 2 × 6 joist will be used. So with a new uniform load on the joist

of 440 pfl, determine the maximum spacing of the cap beams, which is the maximum

span of the 2 × 6.

Some interpolation has to be used on the charts when the uniform loads are in

between the values shown. Always go to the more conservative number. Taking this

into account, cap beams should not be spaced farther than 56′′ OC. If the designer

wanted to make the layout of this system go without the potential of error, 54′′ couldbe selected as an even 4′6′′ OC.

Step 3: Select a cap beam size and determine the safe spacing between the posts.

A 4 × 8 has been selected for a cap beam. The chart selected for this step is the

same as the single-beam chart because a cap beam is a single beam. Be cautious not

to accidentally use the double waler chart as was used in the Chapter 9 examples.

The new uniform load on a single 4 × 8 cap beam is determined as follows:

W = 330 psf × 4′-6′′ = 1481 plf

The single-beam chart indicates that a 4 × 8 cap beam can span 60 in between

posts. Therefore, post spacing should not exceed 60′′ (5′0′′).Step 4: Determine the post load on a single interior post (post design will be

skipped in this example, so refer to Example 10.1).

Pall =Pmax × (cap beam spacing × post spacing)

Pmax

= 330 psf × (4.5′ × 5′) = 7425 lb

Final Design:

3

4′′ BB-OES plywood used in the strong direction

2 × 6 joists spaced at 16′′ OC

4 × 8 cap beams spaced at 4′6′′ OC

Space posts at 5′ 0′′ OC (post size not determined in this example)

Post load = 7425 lb


Example 10.3 Elevated Slab with Dropped BeamsDesign an elevated slab and dropped beam falsework using an aluminum beam and

shoring frame system.

This example will utilize standard form plywood, aluminum joists, and cap beams

supported by 15 k per leg capacity aluminum frames. All components except the

plywood are premanufactured metal products. Appendix 7 will be used for plywood

properties and Appendix 9 will be used for the aluminum beam properties.

The sketch in Figure 10.9 shows a typical section through the slab and dropped

beams. This type of falsework is more complicated because the dropped beams and

the slab are at different elevations. Therefore, two levels of decking are required.

FIGURE 10.9 Slab and dropped beam.

The design load including live load and material will be given as the following:

Elevated slab section = 220 psf

Dropped beam = 480 psf

As mentioned earlier, the aluminum beam charts from the previous chapter will

be sufficient for the joists and cap beams; and each post of the modular frames can

support 15 k, which already includes a 2.2 : 1.0 factor of safety. The frames come in a

variety of widths and heights. Figure 10.10 shows an arrangement of shoring frames

for an elevated slab and beam system.

Step 1:Determine the joist spacing for both the slab and dropped beams using the

same3

4′′ plywood as Example 10.2. Use Appendix 7 for aluminum beam spacing.

Elevated slab = 220 psf, maximum joist spacing = 19′′ OC

Dropped beams = 480 psf, maximum joist spacing = 15′′ OC

Stage 1: Elevated Slab

Step 2A: Determine the maximum spacing of the slab joists using 19′′ spacing and

the chart for aluminum beams in Appendix 9.

W = 220 psf × (19′′∕12′′∕ft) = 348 plf

The aluminum beam chart in Appendix 9 determines the maximum span of the

aluminum joist supporting the elevated slab. It will be assumed that these joists will


FIGURE 10.10 Slab and dropped beam.

span three cap beams (two spans). Therefore, the two-span section of the chart will be

used. In the two-span column, go down to 348 pfl and the beam span will be indicated

on the left, y axis.

Using the 1

4′′ maximum deflection (started values in Appendix 9) the beam can

span 10.5 ft. This is a very long span, but at the same time, the load is very light.

At this point a decision has to be made on whether to space the joists at 10.5 ft or less

to change the cap beam load. To make this decision, list the different cap beam loads

that apply to some various joist spans.

Table 10.2 shows span options for the aluminum beams and cap beams.

TABLE 10.2 Aluminum Beam Spans with Cap Beam Spacing from Appendix 9

Joist Span (ft) Cap Beam Uniform Load (plf) Spans (ea) Cap Beam Spacing (ft)

10.5 2310 2 4.0

9 1980 2 4.5

8 1760 2 5.5

6 1320 2 7.0

5 1100 2 7.5

Another consideration is the size of the slab sections in plan view. If the slab

lengths are 15 ft, for instance, then half of the section would be 7.5 ft. The joists can’t

span the whole 15 ft; therefore, it would make sense to split the distance in half and

span the joists only 7.5 ft. In addition to the geometry of the concrete segments, the

shoring frames come in certain sizes that range from 4-ft dimensions to 7-ft dimen-

sions as shown in Figure 10.11. With this said, it would not make sense for either

span to be more than 7 ft.


FIGURE 10.11 Shoring frame options.

Looking at Table 10.2, many options are available; but if we look at the shoring

frame geometry, the spacing between the frame legs can dictate the beam spacing.

Sometimes in temporary structure design, the beams allow greater spacing than is

practical for the rest of the design. This happens frequently with aluminum compo-

nents because aluminum is so efficient that the rest of the supports can’t match its

strength; therefore, the aluminum gets underutilized to compensate for the system.

This does not mean that it is not economical. Even underutilized, aluminum systems

have great value.

Table 10.3 shows some examples of the available frame sizes that can be purchased

or rented.

TABLE 10.3 Some Available Shoring Frame Sizes

Frame Options

(W) Width

of Frame (ft)

(L) Length withCross Brace (ft)

(H) Heightof Frame

1 4 8 6 ft

2 4 10 6 ft

3 2 8 5 ft

4 2 10 5 ft

1 Extension 4 8 or 10 5′4′′

2 Extension 2 8 or 10 4′4′′

In this example, wewill space the aluminum beams tomatch an efficient aluminum

frame system shown in the charts above. The 4-ft-wide frames are the most common

and should be used whenever space allows. These frames will also maximize the

system by using fewer frames.

Figure 10.12 shows the plan view layout of the shoring frames. The dropped beams

are shown with double lines and the flat slabs are in between.

Stage 2: Drop Beam

Step 2B:Determine the maximum spacing of the drop beam joists using 15′′ spacingand the chart for aluminum beams.

W = 480 psf × (15′′∕12′′∕ft) = 600 plf


FIGURE 10.12 Plan view layout of shoring frames.

The aluminum beam chart in Appendix 9 determines the maximum span of the

aluminum joist supporting the drop beam. It will be assumed that these joists will

span three cap beams (two spans). Therefore, the two-span section of the chart will

be used. In the two-span column, go down to 600 pfl and the beam span will be

indicated on the left, y axis. As before, the span is very long and the geometry of

the structure may ultimately govern the actual beam spans. The chart in Appendix 9

indicates a maximum span of 9 ft using the 1

4′′ max deflection. However, like the slab,

this spacing may not meet the geometry of the structure. The beams are less than 2 ft

wide so only one row of frames per beam will be necessary. If a 4-ft frame is used,

the beam will load a single row of frames. Figure 10.13 shows a single row of frames

under a single cap beam.

FIGURE 10.13 Beam shoring.

Stage 3: Shoring Frames

Step 3: Shoring tower capacities are rated based on what one leg can support with a

1.5–2.5 : 1.0 FOS. The design of the shoring towers involves calculating how much

load (tributary area) is supported by one individual leg. Flat slab reinforced concrete

decks are easily calculated for each shoring leg because the unit weight per square

foot is the same throughout the system. This dropped beam scenario, however, is more

complicated because of the two different thicknesses of the top slab (220 psf) and the

dropped beam (480 psf).


If the top slab shoring creates a 4-ft × 7-ft pattern, the load per shoring leg is as

follows:

220 psf × 7 ft × 4 ft = 6160 lb < 15,000 lb (OK)

Using these same frames under the dropped beam as in Figure 10.13 would result

in the following shoring leg load:

480 psf × 2 ft wide beam = 960 plf of beam

If the same pattern is used (4′ × 7′), then the load for one leg would be as follows.

960 plf∕2 sides × 7 ft = 3360 lb

Or

480 psf × 7 ft = 3360 lb

Most aluminum systems are rated for 10–15 k per leg, which includes the applica-

ble FOS. These shoring towers would be more than acceptable for this example; and

one might even consider a lighter weight steel system (10–11 k).

The case study to follow is an example of how a building contractor decides to

form multiple-story slab decks and the methods used. This case study is purposely

not similar to the chapter examples so diversity can be illustrated.

CASE STUDY: BUILDING FALSEWORK AND

FORMWORK—CHARLES PANKOW BUILDERS, LTD.

Background

Selecting optimal formwork systems for a concrete framed building can be a

daunting task. For the past 50 years, Charles Pankow Builders, Ltd. has been

a leader in the building industry, specifically in buildings that utilize concrete

as the building structure. This case study will provide insights on why Pankow

selected the forming systems they did on two different projects.

Criteria

The following criteria are reviewed by Pankow’s preconstruction group, super-

intendents, and foremen, when selecting form work systems. One of the first

and foremost criteria is selecting forming systems that reduce on-site labor, the

riskiest part of the construction process. Other criteria include:

• Labor productivity

• Familiarity of forming system

• Form reuse


• Prefabrication and rental costs

• Schedule

• Site constraints

In general, Pankow does not own its own formwork and typically rents from

formwork suppliers, which allows the company to be flexible in selecting the

right formwork for the job.

Case Study: 20-Story Apartment Building, Los Angeles

Statistics

• 18 typical residential floors + 2 garage levels

• 8′′ thick mild steel and post-tensioned slabs

• 9′8′′ floor-to-floor height

• 154,000 ft2 per floor, one slab pour per floor

• Rectangular concrete core

• 2 shear walls outside of core

• 15 concrete gravity columns, regularly spaced and in-line

• Hoisting via tower crane (30,000-lb capacity)

For this case study, we will concentrate on the typical floor forming systems

shown in Figures 10.CS3 and 10.CS4.

FIGURE 10.CS3 Twenty-story apartment building, Los Angeles, California

(courtesy of Charles Pankow, Ltd and Doka Forms).

(continued)




(Continued)

FIG

URE10.CS4

Typicalfloorplan

(courtesyofCharlesPankow

,Ltd

andDokaForm

s).


Floor Slab Forming System

Given the size of the floor plate, the regularly spaced and aligned columns,

and the number of reuses of the forms, a column-hung, large fly table system

(rented from Atlas Construction Supply, Inc.) was selected, a system that is

Pankow’s first choice for a project such as this. A fly table slab-forming system

(Figure 10.CS5) provides Pankow good labor productivity, is fast setting, is a

system Pankow’s crews are familiar with, and reduces on-site labor when com-

pared to other systems, for example, handset. The fabrication costs for these

fly form systems are much higher than handset systems; therefore, a high reuse

of fly forms is imperative for these systems to be economical for a project.

A rule of thumb in the Pankow forming “handbook” is that fly tables will make

economic sense when they are regular and can be reused at least 14 times.

FIGURE 10.CS5 Fly table (courtesy of Charles Pankow, Ltd and Doka Forms).

When comparing fly table systems, one aspect to consider is column-hung

supports or truss/shored supports. The advantage of the column-hung systems

over truss/shored systems is they do not require reshores. For truss/shored fly

table systems, reshores are required to be installed three floors below the floor

being placed in order for the wet concrete load to be transferred safely to the

floors below. These shores get in the way of trades working below the concrete

deck. The column-hung fly table system does not require reshores and allows

the drywall and MEP trades to begin layout and installation immediately with-

out reshores being in the way. This helps the overall construction schedule and

(continued)




(Continued)

keeps work flow closely behind the concrete operations. Figure 10.CS6 shows

a fly table being installed.

FIGURE 10.CS6 Fly table installation (courtesy of Charles Pankow, Ltd and Doka

Forms).

Concrete Core Forming System

A crane set, jump form system (Figure 10.CS7) was selected to form the con-

crete core. Because the core walls formed a rectangular box, casting these walls

ahead of the slab was very feasible. Like the flying form systems, jump form

wall systems cast ahead of the slabs have high first costs when compared to

other systems; however, the savings in labor and time makes up for this higher

initial cost.

One convincing reason to cast the core walls ahead of the decks was

schedule. This system allows the core walls and decks to be formed simul-

taneously, thereby reducing the floor-to-floor schedule by at least 3 days,

when compared to casting the walls and decks concurrently. When extended

over the entire duration of the project, this equates to nearly 2 months of

construction time and provides the owner the building early for 2 months of

additional rent.


FIGURE 10.CS7 Jump form system (courtesy of Charles Pankow, Ltd and Doka

Forms).

A self-climbing, jump form system was also evaluated for this project;

however, the project team felt the added cost for the hydraulic system asso-

ciated with a “self-climber” was not worth it. As long as there is available

tower crane time for raising a crane-set jump form system, a self-climber is

not needed. In its spare time, the crane can also service the reinforcing steel

subcontractor.

Also, if the jump form system is nomore than three floors above the working

deck, a personnel hoist does not need to be installed. The jump form system is

accessed via ladders from a platform set inside the core.

Concrete Gravity Columns and Shear Walls

The column and wall forms (Figure 10.CS8) were set off of the previously

cast slab. In order to meet the schedule, a full floor of column and wall forms

were fabricated, and they had to be set and concrete placed in a single day.

All of these forms were stripped the following day. The column forms selected

utilized hinged forms that allowed the forms to be easily set and stripped by the

carpenter crew. All the column and wall forms were flown down to the ground,

requiring adequate storage space. The project site was tight; however, the plaza

level at the ground floor provided enough space for these forms to be stored

during the deck-forming operation.

(continued)




(Continued)

FIGURE 10.CS8 Concrete gravity columns and shear walls (courtesy of Charles

Pankow, Ltd and Doka Forms).

Typical Floor-Floor Schedule (5-day + Sat):

• Day 1: Finish stripping fly tables from floor below, set on column-hung

brackets at floor above (slab was stressed previously).

• Day 2: Install slab rebar and Post Tensioning (PT), set MEP blockouts, set

edgeforms.

• Day 3: Install slab rebar and Post Tensioning (PT), set MEP blockouts; place

concrete core concrete.

• Day 4: Place slab concrete, set column rebar cages, and crane-up column

forms at end of day.

• Day 5: Set column and wall forms, place column, and wall form concrete.

• Day 6: Strip column and wall forms and crane down to ground floor.

NOTE: A minimum of 2 h of overtime (OT) is worked every day to meet

this schedule.

Case Study: 11-Story Apartment Building, San Francisco

Consider the Mission Street apartment building (Figure 10.CS9).


FIGURE 10.CS9 1321 Mission Street apartment building (Courtesy of Charles

Pankow, Ltd and Doka Forms).

Statistics

• 11 typical floors above grade, one garage level

• 6.5′′ thick mild steel and post tensioned slabs

• 9′6′′ floor–floor

• 9000 ft3 per floor, one slab pour per floor

• Nonrectangular (J-shaped) concrete core

• 2 straight shear walls outside of core

• 31 concrete gravity columns, regularly spaced and in-line

• Hoisting via tower crane (13,000-lb capacity)

The typical floor plan is shown in Figure 10.CS10.

Floor Slab Forming System:

A large fly table system is Pankow’s first choice to form these types of slabs;

however, for this project, they did not make economic sense. First of all, the

height of the building only allowed 10 reuses. Second, the shape of the building

and irregular column spacing did not lend itself to large fly tables. And lastly,

high-voltage electrical lines for the Muni bus line on Mission Street did not

allow for the flying of large tables at the lower floors. In this case, Pankow

elected to utilize a “handset” system that resembled “Ellis Shore” style forms

but with a modern twist—Doka’s Doka Flex System (Figure 10.CS11).

(continued)




(Continued)

FIG

URE10.CS1

0Typicalfloorplan.(courtesyofCharlesPankow

,Ltd

andDokaForm

s).


FIGURE 10.CS11 Doka form system (courtesy of Charles Pankow, Ltd and Doka

Forms).

Pankow did evaluate a “mini” table system for this project. An advantage of

these mini tables is the preinstalled perimeter safety rail on the tables. Similar

to the larger fly tables, the perimeter rail is preinstalled at the perimeter and,

therefore, reduces leading edge safety risks. Although the tables have to be

transported to the perimeter for the fly out, this is done very effectively via

an electric transporter, affectionately called the “turtle.” However, in the end,

Pankow elected to go to a handset system because:

The handset system could be utilized from the ground floor slab through the

roof and not require the crews to switch systems to form level 3 slab. The mini

tables could not be utilized to form the ground floor and level 2 slab, a handset

system would need to be used anyway. Switching systems would add another

learning curve for the crews.

The handset system is not as dependent on crane time as the mini tables.

The mini tables would require use of the crane for the entire day they are set

and stripped. The handset system is broken down into small bundles that can

be flown up to the floor above in a short amount of time.

The irregular column layout and shape of the building lent itself to a more

flexible handset system.

(continued)




(Continued)

Many of the crews in the Bay Area are familiar with the Ellis Shore system,

and the Doka Flex System is similar to Ellis Shore, but is quicker and easier

to install. For this reason, the crew’s learning curve with the system would be

shortened.

This system requires reshores 3 floors below the formed slab. However, since

the building is only 11 stories tall, the overall impact of reshores to the schedule

is minimal.

Concrete Core Forming System

Unlike the Los Angeles project case study, the concrete core for the 1321 Mis-

sion project does not lend itself to be cast ahead of the slab. One reason is this

project’s core wall is J shaped and fully enclosed, that is, rectangular. A fully

enclosed jump form “works” better than one that is not fully enclosed. Second,

because there are so many columns (31 each) on the floor and because there

are two additional shear walls outside the main core wall, the time saved by a

jump form is reduced. Third, the added cost of a jump form as compared to a

gang form set off the slab is difficult to amortize over just 11 floors. And lastly,

even if time can be saved with a jump form (say 1 day per floor), that equates

to about 2 weeks on the overall schedule, which is not worth the added cost

(Figures 10.CS12–10.CS14).

The decision Pankow made was to utilize a crane set, gang form system that

is cast concurrently after the slab is cast. This system requires the forms to

be flown to the ground while the decks are being formed, and on this site, the

storage area was very limited. In order to decrease the required storage space,

the forms on the inside of the elevator were set on a platform covering the

elevator shaft and did not need to be flown to ground.

Concrete Gravity Columns and Shear Walls

Pankow selected similar forming systems for these elements as the Los Ange-

les project. Hinged column forms and gang wall forms (Figure 10.CS15) were

used. Due to the large number of columns and walls, a 3-day schedule was

established for the forming and casting of these elements.

Typical Floor–Floor Schedule (8-day with Sat)

• Day 1: Strip deck forming system from floor below (slab was stressed pre-

viously), set on slab above.

• Day 2: Finish setting deck forming system, install slab rebar and PT, setMEP

blockouts, set edge forms.


FIGURE 10.CS12 Column forms (courtesy of Charles Pankow, Ltd and Doka Forms).

(continued)




(Continued)

FIG

URE10.CS1

3Jwall(courtesyofCharlesPankow

,Ltd

andDokaForm

s).


FIGURE 10.CS14 Column forms (courtesy of Charles Pankow, Ltd and Doka

Forms).

FIGURE 10.CS15 Wall forms (courtesy of Charles Pankow, Ltd and Doka Forms).

(continued)




(Continued)

• Day 3: Install slab rebar and PT, set MEP blockouts.

• Day 4: Install slab rebar and PT, set MEP blockouts, set column and wall

rebar cages, raise core wall platform.

• Day 5: Place slab concrete, set column andwall rebar cages, crane-up column

forms at end of day.

• Day 6: Set column and wall forms on previously placed slabs and strip edge

form (reduced crew size).

• Day 7: Pour1

2of columns, pour

1

2of shear walls, set wall forms, Stress slab

PT.

• Day 8: Strip and set columns, pour second1

2of columns, pour

1

2of shear

walls.

• Day 9: Strip columns and walls, hoist the forms to the ground.

10.1.6 Bridge Project

Your professor will go over the design of a basic box girder falsework system.

Box-Girder Falsework Design A box-girder bridge is a very common bridgedesign used by various county, state, and federal agencies and is predominantly rein-forced concrete. The foundations are the same as any other type of bridge using drivenor drilled piles, concrete footings, and concrete piers. The cap beams spanning thepiers are typically incorporated with the superstructure. In other words, the cap beamsare within the depths of the superstructure and not between the superstructure andthe top of the piers. The superstructure is a series of open boxes consisting of a topslab (driving surface), vertical stem walls, and a bottom slab. The number of boxesdepends on the width of the bridge and the height of the box depends on the bridgespan lengths. The longer the spans, the deeper the bridge is (more moment of iner-tia). The stem walls contain post-tensioning ducts with high-strength stands that aretensioned after the top deck is placed and cured.

The design of a box-girder falsework system is usually very complicated.However, the design can be simplified if all the conditions are reduced to basicscenarios such as:

Simple box-girder geometry

Not spanning over water, live traffic or environmentally sensitive areas

Heights less than 30 ft

If these conditions are favorable, then a fairly simple step-by-step list can be devel-oped in order to produce an economical falsework system. Figure 10.14 shows a verycommon, but simple box-girder bridge cross section.


FIGURE 10.14 Box-girder bridge cross section.

Step 1: Stringer loadsAssume a stringer will be at each location A through F line (6 stringers). The

two side stringers (A and F) are under the edge of deck (EOD) and the rest of the

stringers are under the centerline of each outside and inside stem wall (B, C, D, and

E). Determine the associated concrete area of each line. For example, stringer B will

have to support half the overhang to line A and half the box to line C. Therefore, line

B will take a shape like Figure 10.15.

Draw each section A–F and calculate the concrete volume (CF) of that section.

Convert the volume into a weight using 160 pcf. Consider this section as a 1 ft-long

bridge section and multiply by 1 ft.

Weight = Area × 160 pcf × 1 ft

Add a live load of 50 psf over the shadow area of the section.

Add 100–150 plf of bridge for the self weight of the stringer, which is not

known yet.

Step 2: Soffit DesignFind the widest box width (stem wall to stem wall) and calculate the linear load

from the bottom slab. Assume plywood and a joist has to span this distance and sup-

port the bottom slab load. Figure 10.16 shows this section.

W = (thickness∕12′′∕ft) × 160 pcf × 1 ft

Add live load of 50 psf for live load.

Use the charts in Appendix 7 to determine the best plywood and joist size based

on this load.

FIGURE 10.15 Stringer line B tributary area.


FIGURE 10.16 Bottom slab spanning stem to stem.

Step 3: Stringer DesignDetermine the longest span of the bridge. Using the uniform loads per linear foot

of bridge for each stringer line, determine the minimum size of each stringer A–F.For example, if line C and D weigh 3000 plf of bridge and the longest span is 34 ft,the minimum stringer moment and section modulus would be as follows:

M = (3000 lb)(34′)2∕8 = 433,500 ft-lb

Sx = (433,500 ft-lb)(12′′∕ft)∕21,600 psi = 240.8 in3

Perform this calculation for all lines that are different. Notice that A and F are thesame, B and E are the same, and C and D are the same. Therefore only three of thesecalculations should be run. A table like Table 10.4 should be used in order to keepthe work organized. Only C and D are filled in using the w = 3000 plf and L = 34 ft.

From this information in the table, determine howmany braces each stringer wouldneed and decide which of the three stringers are best for each line with a 34-ft span.

Step 4: Top Cap DesignDetermine the minimum size of the top cap beam supporting the stringers. Select

how many posts will be needed. In the bridge section used in this example, threeto four posts would be adequate. Using the linear loads from A–F and the 34-ftspan, apply each total to the top cap beam and set three to four posts as supports.Figure 10.17 shows this FBD.

This is an indeterminate structure with four posts (supports). Using a beam pro-gram would be the most accurate and less complicated. Whichever method is used,determine the lightest top cap beam that will support this loading condition. The loadsare set by the stringer layout and cannot be changed. However, the posts can bemovedleft and right to establish the lowest moment in the beam, thus, economizing thedesign. Finally check for web yielding over each post and determine if stiffeners arerequired.

TABLE 10.4 Stringer Design

Lines A and F Lines B and E Lines C and D

Linear load (W = plf) ? ? 3,000

Length (ft) 34 34 34

Mmax (ft-lb) ? ? 433,500

Minimum section (Sx) ? ? 240.8

Beam option 1 ? ? W18 × 130



Lateral braces required (#) ? ? lb = 20,000∕Fy(d∕Af )


FIGURE 10.17 Stringers loading top cap beam.

Step 5: Post DesignDetermine the load on the heaviest loaded post using that line’s longest span and

heaviest linear load. Assume that lines C and D have the heaviest linear load and the

span of 34 ft is the longest.

P = W × L

P = 3000 pfl × 34 ft = 102,000 lb (102 k)

Now determine the longest post length. In order to do this, elevations from the

drawings are used and the elevation from the top of bridge to the ground needs to be

known. The post length then becomes as follows:

Bridge top deck elevation (elev) − Gound elevation (elev)

= gross dimension (ft) − decking material (ft)

− tallest stringerand top cap (ft)

− pads, corbels,wedges, and sandjacks and bottom cap (ft)

= post length (ft)

With the maximum post load (P) and the maximum post length (L) calculate theminimum (square) post required:

P = (0.3)(E)(l∕d)2

× area ≤ Pact

Step 6: Lateral bracing load of the heaviest loaded falsework bent.

Determine the total bridge weight at the heaviest loaded bent. This would be the

P1,P2 and P3 loads all added up from step 4.

F = (2 × P1) + (2 × P2) + (2 × P3)

Apply 2% (minimum load) laterally to the top cap beam of the bent as shown in

Figure 10.18.

Step 7: Stem wall formwork design.

Design the wall formwork for one of the internal stemwalls (lines C or D). Assume

full liquid head pressure for P, 58′′ plywood, all 2 × 4 studs and walers, and snap ties


FIGURE 10.18 Lateral force.

with max capacity of 3500 lb. The wall should be formed from the top of the bottom

slab to the bottom of the top slab. In other words, the wall height is equal to the bridge

thickness minus the top and bottom slab.

Step 8: Top Deck Formwork Design

Design the formwork for the top deck slab from stem wall B to C or C to D,

whichever length is greater. By choosing the widest box, the worst-case scenario is

used. Use5

8′′ plywood, 2 × joists, 4 × cap beams, and 4 × 4 posts.

CHAPTER 11

BRACING AND GUYING

Guying and bracing of temporary structures has become a much talked about and

scrutinized subject in construction. The failure of such systems can not only be catas-

trophic but can prolong the misperception that construction is an unsafe industry.

In addition, guying and bracing is a very visible aspect of construction; it is often

what the public sees daily on their commute to work and hears about when accidents

happen. The case study accident in Chapter 10 is an example of a guying and bracing

accident. It was placed in Chapter 10 because it involved falsework.

Tilt-up construction, formwork placement, and rebar cage guying and bracing

operations have indeed resulted in several high-profile accidents in the last 20 years.

Iron workers, carpenters, laborers, and operators are among the tradesmen who have

been severely injured or killed in these accidents. When these accidents occur, they

are usually not “reportable” OSHA cases; they are more often “recordable,” “lost-

time,” and, unfortunately, “fatality” OSHA cases. Reportable accidents are accidents

that occur and do not require a doctor or prescription medication. A recordable acci-

dent, however, is an accident that does require medical attention and/or prescription

medication. A lost-time accident means the injured person misses his next scheduled

day of work.

These injuries and fatalities are not always caused by the design of the actual

supports, but many times are the result of human mistakes and bad judgment. It is

therefore crucial that designers and builders use the utmost caution when provid-

ing support for their temporary structures. Figure 11.1 shows a pier table form and

reinforcing steel being supported by custom pipe braces.

267

268 BRACING AND GUYING

FIGURE 11.1 Custom pipe bracing.

11.1 REBAR BRACING AND GUYING

Rebar bracing and guying are operations more likely to fail than others mentioned

above. The reasons are multiple, but the size, weight, and the inherent unstable char-

acteristics of rebar cages are primary reasons. Rebar is typically connected by tie

wire, and the engineer of record for the permanent structure does not always take

into account the tripping (standing), supporting, and guying of the vertical, unsta-

ble members. This responsibility is borne by the contractor and his engineer, who

typically works for a different firm than the designer of record for the permanent

structure. State agencies have begun to issue more guidelines for the contractor to

stabilize rebar assemblages. For instance, the specifications on bar reinforcing cages

measuring 4 ft in diameter or greater may have the following requirements:

A minimum number of bars, depending on the bar size, of each cage, equally

spaced around the circumference, shall be tied to a minimum number of rein-

forcement intersections by double tie wire. Theminimumnumber of bars would

be selected by the designer and approved by the owner/agency.

A minimum percentage of remaining intersections may have to be tied with single

tie wire connections. Staggering may be required.

Internal bracingmay be required to avoid collapse during assembly, transportation,

and hoisting.

Such requirements are becoming the standard for state and federal highway

projects. It should be only a matter of time before the rest of the construction

industry adopts such practices.

11.2 FORM BRACING WITH STEEL PIPE AND CONCRETE DEADMEN 269

Rebar guying, if properly planned, does not have to be limited to external guycables or stiff-leg bracing. Internal bracing can be added to the inside of the unstablecages that add rigidity during the lifting and guying stages. This text will not provideexamples of this internal bracing, but it should be noted that this practice will lessenthe stresses on the tie wire and the vertical and hoop rebar during the different phasesof cage guying. In addition, in most cases, the planning and engineering that goesinto the bracing of the rebar cage also translates to a form guying and bracing planas well.

The first example in this chapter will cover a wall form or panel being braced toresist a typical wind force. The wall panel will be braced with a pipe brace in tensionand compression, and a concrete anchor block called a deadman. Typical pipe bracesconsist of a smaller pipe inside a larger pipe and is designed to the smaller pipe.The ends of the pipe braces usually have adjustment screw jacks and plate steel forattachment. The wind chart from Chapter 3 has been included in this chapter as wellfor quick access to loading forces converted from mph to psf. The second examplewill be geared toward highway, bridge work, and reinforcing steel cages with wirerope guying systems.

11.2 FORM BRACING WITH STEEL PIPE AND CONCRETE DEADMEN

Example 11.1 Wall Form/PanelDetermine the loading on a pipe wall brace from the wind forces given, then size the

pipe braces and the deadman (concrete anchor block) including the required factors of

safety of 1.5 : 1.0. The wind force is 80 mph and the braces and deadmen are at 15-ft

centers. Design the braces for tension and compression, and design the anchor blocks

to resist sliding in both directions as well. Use Figure 11.2 for additional information

and dimensions.

FIGURE 11.2 Example 11.1 sketch.

Steps to designing a wall panel bracing system:

1. Calculate the wind force on a 12′′ strip of wall panel.

2. Distribute the force to a single brace based on the brace spacing and convert to

force to a single resultant force (R).


3. With statics, and using the slope of the brace, determine the force in the brace

and anchor block.

4. Determine the pipe brace length and size the pipe as a single pipe (smallest,

inside pipe) to resist buckling (compression) and tension.

5. Transfer the brace loads to the anchor block and determine the minimum block

weight to resist sliding.

Table 11.1 can be used for the forces generated from local wind speeds. Lets

assume a 20-ft wall with wind speed of 80 mph.

TABLE 11.1 Wind Speed and Force

Speed (mph) Force (psf)

10 0.27

20 1.08

30 2.43

40 4.32

50 6.8

60 9.7

70 13.2

80 17.3

90 21.9

100 27.0

110 33

120 39

130 46

140 53

150 61

Pipe braces at 15-ft OC attached to concrete deadmen are shown in Figure 11.3.

FIGURE 11.3 Wind distributed load.

Step 1: Determine the force from the wind as a 12′′ strip load. From the chart, use

the loading that corresponds to an 80-mph wind force. Table 11.1 indicates that an

80-mph wind produces 17.3 psf force on the wall panel. If the panel was higher than

30 ft, an additional 5 psf would be added because wind forces are greater the higher


the obstruction. This will be discussed more in Example 11.2. The load on a 12′′ stripis as shown here.

R (12′′ strip) = 17.3 psf × 20 ft = 346 per linear foot of wall

Step 2: Distribute the force to one brace. Since each brace supports 15 linear feetof wall, the 12′′ strip load has to be multiplied by the 15-ft brace spacing.

R (one brace) = 346 plf of wall × 15 ft of wall = 5190 lb per 15 ft of wall.

Figure 11.4 illustrates this FBD.

FIGURE 11.4 FBD of brace connection to back of form.

Step 3: Determine the force at the brace connection to the back of the wall panel

(location B). The resultant force of this 20-ft strip load is 10 ft from the bottom

because the resultant location for a rectangular load is one-half the distance of the

rectangle. Apply this 5190 lb force 10 ft from the base of the wall and determine the

reaction at the pipe brace connection toward the top of the wall. The brace is attached

3 ft from the top of the wall (17 ft from ground), the brace is sloped at 2V : 3H, and

the deadman is 3 ft tall. The base of the panel (ground) is location A.

The horizontal component is FHB, the vertical component is FVB, and the axial

force through the pipe brace (2 : 3 slope) is F. Summing the moments about the base

of the panel will determine one unknown: the horizontal force at B. Since FVA is in

line with the connection point, there is no moment created by this force. Also the base

of the panel is in line with point A as well, so again there is no moment generated

from its reaction.

ΣMA = 0

5190 lb × 10′ = FHB × 17′, FHB = 3053 lb

Figure 11.5 shows the FHB at location B.

FIGURE 11.5 FBD at location B.


Using the 2V : 3H slope determine FVB:

2

3=

FVB

3053FVB = 2035 lb

The axial force (F), therefore, is

F =√(3053)2 + (2035)2 = 3669 lb

The horizontal and vertical forces can be used to design the connection at the

back of the panel. This will not be done in this example; but on a real design, the

connection would have to resist these two forces for shear and tension on all members

involved with the connection. The axial force will be used in the next step to design

the pipe brace.

Step 4: Design the pipe brace for buckling and tension. Buckling is typically the

weak link, so the brace will be designed for buckling first and then tension can be

checked.

As mentioned in Chapter 2, the following is the calculation for allowable buckling

stress. The factor of safety used in this example is 1.5 : 1.0.

Fbs =𝜋2 E

(kLe∕r)2 × FOS= PA


E = modulus of elasticity for steel, usually 29,000,000 (psi)

k = connection type multiplier, depending on fixed, hinged, etc., the

connection to the back of the panel and deadman is hinged (1.0)

Le = effective length of column, if there are supports, then the longest

distance between supports (in), this pipe brace is not supported in the

center

r = weakest radius of gyration (in). For pipe, there is only one radius of

gyration due to its circular shape.

Refer to the pipe chart in Appendix 2 for pipe properties up to 12 in in diameter.

P∕A is normal stress and is set equal to allowable buckling stress in order to deter-

mine (P)all or (A)rea allowed.The trial-and-error method can be used. First determine the length of the pipe with

the dimensions shown in Figure 11.6. Using the 2V : 3H slope and a vertical distance

of 14 ft (20′–3′top–3′ btm), calculate the horizontal distance2

14

′= 3∕H = 21 ft.

FIGURE 11.6 Pipe brace slope.


Now using the Pythagorean theorem, figure the pipe length on the diagonal:√(14′)2 + (21′)2 = 25.24 ft

A 31

2′′ standard pipe will be tested first. Look up the cross-sectional area (A)

and the radius of gyration (r) for this pipe in Appendix 2.

A = 2.68 in2

r = 1.34 in

Calculate Fbs

Fbs =𝜋2(29,000,000)

{[1.0(25.24 × 12∕ft)∕1.34]2} × 1.5= 3735 psi

Set Fbs = Pall∕A and solve for Pall = 3735 psi × 2.68 in2 = 10,009 lb

This 25.24-ft-long, 31

2′′ standard pipe can support 10,009 lb before failing in

buckling. If buckling was not an issue, then Pall = 27,000 psi × 2.68 in2 = 72,260 lb

tensile strength. This is a 62,251-lb reduction because of buckling.

The required load in the pipe brace was only 3669 lb. There is already a factor of

safety of 1.5 built into our allowable load of 10,009 lb; so this means if we use this

pipe, the actual FOS is 10,009∕(3735∕1.5) = 4.02. This is very high, so a smaller

pipe should be considered.

Try a 3′′ standard pipe. Look up the cross-sectional area (A) and the radius of

gyration (r) for this pipe:

A = 2.23 in2

r = 1.16 in

Calculate Fbs:

Fbs =𝜋2(29,000,000)

[(1.0(25.24 × 12′′∕ft)∕1.16′′)2] × 1.5= 2799 psi

Set Fbs = Pall∕A and solve for Pall = 2799 psi × 2.23 in2 = 6242 lb > 3,669 lb

(OK).A 21

2′′ pipe was tried by the author and could only support 3171 lb, so the 3′′ pipe

is good with a little more FOS built in.

Check the pipe in tension: 3669 lb∕2.23 in2 = 1645 psi < 27,000 psi (OK).Step 5: Transfer the horizontal and vertical pipe brace loads down the deadman

anchor block and determine the minimum weight of the deadman.

In physics, friction was taught. At that time, the future construction management

student probably never dreamed that this would come in handy in his field of study.

Now is the chance to refresh one’s memory of physics on how friction works.

Figure 11.7 shows the pipe brace forces being transferred to the deadman anchor

block. The same diagonal, vertical, and horizontal forces still apply.


FIGURE 11.7 FBD at deadman anchor block.

Contrary to what most people believe or what common sense may appear to say,

friction is not based on the surface area at the bottom of the deadman. Friction gets

its resistance from the normal force and the coefficient of friction value associated

with the two surfaces involved, the bottom of the concrete block and the ground.

Depending on the construction site, the ground can be compacted soil, uncompacted

soil, compacted base rock, asphalt, concrete, and the like. When the surface is

unknown, always use a conservative coefficient, the lowest value. Table 11.2 shows

some estimated friction values used in construction. The ranges used typically mean

the difference between wet and dry surfaces. The wet surface would always be the

lower coefficient value.

TABLE 11.2 Coefficient of Friction Values for Standard

Materials in Construction

Surface 1 Surface 2 Coefficient (𝜇)

Concrete Steel 0.35–0.45

Concrete Loose soil 0.25–0.35

Concrete Concrete (rough) 0.50–0.65

Concrete Clay 0.20–0.40

Concrete Sand 0.40–0.60

Concrete Gravel 0.50–0.60

Concrete Rock 0.50–0.70

The deadman size needs to be checked for the pipe in tension and in compression.

When the pipe is in compression (which is how the pipe was sized), the deadman is

being pushed down and to the right. When the pipe is in tension, the deadman is being

pulled and lifted up to the left. It does not matter that the pipe was OK in tension. The

fact of the matter is if the wind blows right to left, the pipe brace will still pull and

lift the deadman.

Physics taught us that friction force is equal to the normal force (N) times the

coefficient of friction (𝜇) mentioned above and shown in Figure 11.8. This force is

a horizontal direction (x) force and the normal force is a vertical direction (y) force.The forces coming from the pipe brace can be converted back into an x and y forceso that all the forces are either horizontal or vertical.

FHB = horizontal brace component (−x when pulling left, +x when pushing to

right)

FVB = vertical brace component (−y when downward, +y when upward)


FIGURE 11.8 FBD push and pull friction forces.

W = weight of deadman vertically downward (−y, downward only)

N = normal force vertically upward (+y upward only)

F = horizontal force (N𝜇), (−x when resisting left, +x when resisting right)

Draw two free-body diagrams, one representing the downward force of the pipe

brace on the deadman and the other representing the upward force of the pipe brace

on the deadman. From these free-body diagrams, total the forces in both the x and ydirection in order to solve for the unknowns.

Using Table 11.2, a coefficient of friction of 0.35 is estimated since the project

conditions will be different for each wall location. Sometimes the deadman will be

on compacted aggregate base and sometimes on soil. A lower number would be more

conservative and require a heavier deadman. There are five variables listed above.

Friction is based on the normal force (unknown) and the coefficient (known). This

means there are really only four variables; two are known (from the pipe brace), and

two are unknown. Figure 11.9 shows anchor blocks being used as a support system

for bridge falsework.

Do we add the forces in the x or y first? Looking at the FBDs, it is apparent

that, besides the known forces, there are three forces vertically in the y and only twoforces horizontally in the x. Therefore, it would make sense to add the forces where

there are fewer unknowns and then use this information to figure out the remaining

forces. From the pipe brace there is FHA = 3053 lb and FVA = 2035 lb. A factor of

safety also needs to be considered so a 1.5 : 1.0 will be included in the following

illustration.

ΣFx = 0 and first solve for the normal force (N). FHA × FOS − 𝜇N = 0

(+3053 lb × 1.5) – 0.35N = 0 N = 13,084 lb

Now we have solved one of the two unknowns (N = 13,084 lb) and it already

includes an FOS. The forces can now be added in the y direction with a value for N.ΣFy = 0, using the new value for N. Remember, N has the FOS built in so do not

apply another FOS.

–FVA −W + N = 0

− 2035 lb –W + 13,084 lb = 0 W = 11,049 lb


FIGURE 11.9 Falsework bent supported by pipe braces.

The deadman weight has been determined when the pipe is pushing downward,

wind blowing left to right. Now, reverse the direction of the pipe brace force, wind

blowing right to left and the pipe brace lifting and pulling in this same direction. The

normal force will be the same because, even though the direction reversed, they all

reversed; therefore, the positive and negative signs reverse, but the magnitude of the

normal force is the same value.

ΣFx = 0 and first solve for the normal force (N). FHA × FOS − ¯N = 0

(+3053 lb × 1.5) – 0.35N = 0 N = 13,084 lb

The forces can now be added in the y direction with a value for N.

ΣFy = 0, remember, N has the FOS built in so do not apply another FOS.

+ FVA −W + N = 0

+ 2035 lb –W + 13,084 lb = 0 W = 15,119 lb

The deadman in this case is much larger (Figure 11.10). You can see by the sign

change of the FVA, the 2035-lb value was now added to the normal force instead of

subtracted. This increases the deadman requirement by 2 × 2035 lb.

How big does a concrete anchor block have to be to weigh 15,119 lb? Concrete

weighs approximately 150 pcf. With this information, we can determine the volume

required for a deadman to weigh 15,119 lb.


FIGURE 11.10 Tipping analysis.

15,119 lb = (b × w × h) 150 pcf. If the dimensions of the deadman were all the

same, say c, we could take the cube root of the volume:

c = 3

√15,119

150= 4.65 ft

The volume of this deadman needs to be 100.8 ft3 if it is made of 150 pcf concrete.

What if the base of the deadman was set at 5 ft × 5 ft and we needed to know the

height (h)?

h = 100.8 ft3

5 ft × 5 ft= 4.03 ft high

Here are two deadman sizes that would work. However, will they tip over if the

pipe brace is connected to the top? Let’s try the first deadman since it is taller and

more susceptible to tipping.

The upward forces will be used since that is how the deadman was sized. There is

a deadman weight acting downward of 15,119 lb and in the middle of the deadman

(1

2of 4.65 ft). There is FVA acting upward with 2035 lb in the middle of the dead-

man ( 12of 4.65 ft). Finally, there is FHA acting to the left with 3053 lb at the top of the

deadman, which is 4.65 ft high. If the moments are added about the bottom left corner

of the deadman and the weight (W) is left unknown, the stability can be determined.

Clockwise will be negative and counterclockwise will be positive.

ΣMA = 0 + (3053 × 4.65) – (W × 2.33′) + (2035 × 2.33) = 0

W = 8128 lb < 15,119 lb (OK)

The minimum weight of the deadman so that it does not tip over is 8128 lb. The

actual deadman weight based on sliding has to be 15,119 lb. Therefore, the deadman

will not tip over. Figure 11.11 shows pier forms guyed with a wire rope cable inside

a highway median. Sometimes the real estate to place anchor blocks is limited and

K-rail is used. However, the weight and geometry of the K-rail must be analyzed in

the same way the anchor blocks were designed.

The attachment of the pipe brace to the top of the anchor blocks should be

designed. The horizontal force can be used to check the shear of the connection. The

vertical, uplift force can be used for the connection in tension. Both of these must be

satisfied. This analysis will be performed in Example 11.2.


FIGURE 11.11 Pier forms guyed with wire rope.

11.2.1 Life Application of Friction Forces

For the student who likes real-life examples to help understand certain concepts, an

illustration from the gym will be used. This example shows a trainer in Figures 11.12

and 11.13 exercising by pushing and pulling a weight inside a metal wheelbarrow

tub—without the wheels, of course.

FIGURE 11.12 Pushing the sled.

11.3 REBAR GUYING ON HIGHWAY PROJECTS 279

FIGURE 11.13 Pulling the sled.

In the gym, pushing or pulling a steel wheelbarrow tub full of weight on

rubber mats can give a person an appreciation for friction forces. In the photos,

the trainer is shown pushing and pulling the metal tub in two different directions.In Figure 11.12, the trainer is pushing the tub. Notice the angle of the trainer’s arms.

Like Example 11.1, the angle of the pipe brace can greatly affect the resistance ofthe deadman as the wheelbarrow will with the trainer. The rubber mats are going to

have a different coefficient of friction with steel then the deadman had with the soilor the rock.

In Figure 11.13, the trainer is pulling the tub. This is the same as the pipe bracepulling and lifting the deadman. In Example 11.1, the pulling and lifting of the dead-

man was what determined the minimum weight of the deadman. So, it would makesense that it is easier for the trainer to pull the tub than push the tub if a correla-

tion is made between the two examples. The coefficient of friction between steel andrubber is about 0.60–0.70. Therefore, the same forces in both examples would result

in greater forces from the pipe brace or the trainer to move the wheelbarrow tub

(deadman).

11.3 REBAR GUYING ON HIGHWAY PROJECTS

As mentioned at the beginning of this chapter, guying and bracing rebar cages onhighway projects is risky and dangerous. Recently, in 2013, the Construction Institute

of the American Society of Civil Engineers sponsored a “Rebar Cage Constructionand Safety—Best Practices” study. The study is one of the first to address the specific

concern of rebar cage stability during construction. During the writing of this book,the author toured two projects in northern California that had many challenging rebar

guying conditions.


The problem and concern with guying tall, unstable rebar cages has always been

an issue; however, in light of accidents that have occurred in the past 15–20 years,

measures have been taken by state and federal agencies, owners, engineers, and con-

tractors to eliminate accidents and even near-misses. The following text and example

is modeled from an actual project in northern California with state jurisdiction and

oversight. The design calculation includes current factors of safety used by engineers

and state and federal agencies. The designs on this and similar projects are reviewed

and approved by at least two separate engineers, either employed by the contrac-

tor and/or the owner/engineer. If this strict requirement of reviews is not specified by

the contractor, then they should be at least specified by the owner. Hopefully, they are

standard practice for both parties.

Example 11.2 The guying of a 40-ft-tall × 10-ft-wide rebar cage and pier form is

shown in Figure 11.14.

FIGURE 11.14 Example guying sketch.

Step 1: Determine all the wind forces and the reactions at the cable locations.

In California and other state and federal projects, the forces on vertical obstructions

such as bridge piers are dictated by the data shown in Table 11.3 and is based on the

height of the structure. As the table shows, the minimum force used in determining

the guying and bracing system is 20 psf at 0–30 ft in height above grade. As the

structure gets taller, the design force increases, as shown in Table 11.3. Now it would

make sense that if the construction was taking place in an area that had more than

TABLE 11.3 Wind Pressures Based on Height above Grade

Height Zone (height above grade in feet) Wind Pressure (psf)

0–30 20

31–50 25

51–100 30

Over 100 35

Source: CalTrans Falsework Manual.


normal wind forces, the pressures used would represent the local conditions and the

values would scale up from the height above grade.

In this example, the height of the column is 40 ft. One interpretation would be to

apply 25 psf from 0–40 ft as a uniform load. The intent of Table 11.3, however, is

to apply the 0–30 ft pressure (20 psf) as a rectangular load from the bottom of the

form to the 30-ft elevation and then 25 psf across the top 10 ft. Once the pressure is

determined, the location of the guy cables should be decided. Sometimes, in order to

decide on one level of cables versus two or three levels of cables, the designer should

run some simple resultant calculations in order to see if a smaller number of guys

will be too much for one level. For this example, it has been decided to place a set of

four guys 12 ft from the ground and place a set of four guys 36 ft from the ground.

The guys will be placed at a 45∘ angle. With this information, the pressure diagram

will look like Figure 11.14. The widest form face is 10 ft.

Calculate W1 and W2 if the widest form face is 10 ft.

W1 = 20 psf × 30′ × 10′ = 6000 lb

W2 = 25 psf × 10′ × 10′ = 2500 lb

Point A is located at the bottom of the form at ground level, so there is a point to

sum moments. Because two levels of cables were selected, there are two unknowns.

This is referred to as an indeterminate loading condition and sum algebra will be used

to determine R1 and R2. Below this calculation, a computer program will be used to

check the resultant values:

ΣMA = 0

(W1 × 15′) + (W2 × 35′) – (R1 × 12′) – (R2 × 36′) = 0

ΣFx = 0

W1 +W2 = R1 + R2, W1 +W2 = 6000 lb + 2500 lb = 8500 lb

therefore R2 = 8500 lb − R1.

Now go back to summing the moments about point A and plug R2 in as 8500 − R1.(6000 × 15′) + (2500 × 35′) = 12′ R1 + 36′ (8500 − R1), with algebra, solve

for R1.90,000 + 87,500 = 12′ R1 + 306,000 − 36′ R1, 90,000 + 87,500 − 306,000 = 12′

R1 − 36′ R1

R1 = 128,500∕24′ = 5354 lb

Finally, R2 = 8500 lb − 5354 lb = 3,146 lb

At this point, W1, W2, R1, and R2 have been solved as values in pounds of force

or reactions.

The computer-generated force diagram in Figure 11.15 confirms the values for R1

and R2 at 3.15 k and 5.35 k, respectively.

The arrows in Figure 11.15 represent theW forces from the wind, and the supports

indicate the reactions where the cables are connected. The cables are connected at 45∘angles, so these horizontal forces will be converted to vertical and diagonal forces in

order to size the wire rope guy cables and the weight of the anchor blocks.


FIGURE 11.15 Computer force diagram.

Step 2: Determine the minimum requirement for the wire rope size of the lower

and upper guy cables. Draw FBD similar to Figure 11.16.

FIGURE 11.16 FBD of upper and lower guy cables.

Lower Guy Cables

FL = FH or FV × 1.414 (45∘ angle) = 5354 lb × 1.414 = 7571 lb

Upper Guy Cables

FU = FH or FV × 1.414 (45∘ angle) = 3146 lb × 1.414 = 4448 lb

FL and FU represent the lower and upper cable forces. Depending on the FOS

used, this will determine the SWL of the cable and, ultimately, the size of the cable.

Table 11.4 shows wire rope breaking strength and the SWL depending on the FOS

chosen by the engineer. Earlier in this book, it was implied that a 5.0 or 6.0 : 1.0 FOS

should be used on all wire rope. In this case, it is up to the discretion of the design

engineer and the governing agency specifications. The wire rope charts in these tables

have been developed from similar IWRC charts, and averages were used to represent

several different manufacturers of wire rope capacities.

Wire rope clips are used to join wire rope back to itself when creating a loop at

the end connection point. The wire rope industry has determined that when these

clips are used, the capacity of the wire rope can be compromised. For that reason, the

minimum breaking strength and allowable loads are reduced to 80 or 90% of the wire

rope capacity. Table 11.5 lists the same information except includes the reduction of


TABLE 11.4 Estimated Allowable Wire Rope Capacitiesa

Minimum Breaking Force and Allowable Loads (tons) for 6× 19 IWRC-XIPS Wire Rope

Wire Rope

Diameter

(in)

Breaking

Strength

(tons)

SWL with

2.0 FOS

(tons)

SWL with

3.0 FOS

(tons)

SWL with

4.0 FOS

(tons)

SWL with

5.0 FOS

(tons)

SWL with

6.0 FOS

(tons)

3∕8 7.6 3.8 2.5 1.9 1.5 1.31∕2 13.3 6.7 4.4 3.3 2.7 2.25∕8 20.6 10.3 6.9 5.2 4.1 3.43∕4 29.4 14.7 9.8 7.4 5.9 4.97∕8 39.8 19.9 13.3 10.0 8.0 6.6

1 51.7 25.9 17.2 12.9 10.3 8.6

aSWL = safe working load, IWRC

TABLE 11.5 Estimated Allowable Wire Rope Capacitiesa

Minimum Breaking Force and Allowable Loads for 6× 19 IWRC-XIPS Wire Rope with

Wire Rope Reduction of 80% < 1′′ and 90% 1′′ and Greater

Wire Rope

Diameter

(in)

Breaking

Strength

(tons)

SWL w/2.0

FOS &

80–90%

SWL w/3.0

FOS &

80–90%

SWL w/4.0

FOS &

80–90%

SWL w/5.0

FOS &

80–90%

SWL w/6.0

FOS &

80–90%

3∕8–80% 7.6 3.0 2.0 1.5 1.2 1.01/2–80% 13.3 5.3 3.5 2.7 2.1 1.85∕8–80% 20.6 8.2 5.5 4.1 3.3 2.73/4–80% 29.4 11.8 7.8 5.9 4.7 3.97∕8–80% 39.8 15.9 10.6 8.0 6.4 5.3

1–90% 51.7 23.3 15.5 11.6 9.3 7.8

aSWL = safe working load, IWRC.

wire rope capacity to 80 or 90% when these clips are used. Figures 11.17 and 11.18

show a wire rope clip and how they are used to make connections, respectively.

Using an FOS of 4:1, the wire rope size required by Table 11.5 would be:

Lower 7571 lb/2000 lb/T = 3.8T, use 5/8′′ wire rope (SWL = 5.2T > 3.8T, butabove 1

2′′ 3.3T).

Upper 4448 lb/2000 lb/T = 2.2T, use 1

2′′ wire rope (SWL = 3.3T > 2.2T, but above

3

8′′ 1.9T).Using an FOS of 4 :1, the wire rope size required by Table 11.6 would be as follows

per level:

Lower 7571 lb/2000 lb/T = 3.8T, use 5


1

2′′ 2.7T).


FIGURE 11.17 Wire rope clip.

FIGURE 11.18 Wire rope connection with clips and thimbles.


FIGURE 11.19 Pier guying system.

Upper 4448 lb/2000 lb/T = 2.2T, use 1


3

8′′ 1.5T).Figure 11.19 is a photo of a pier form being guyed with wire rope and deadmen

anchors.

The wire rope selection shows that the reduction factor for the use of wire rope

clips did not affect the size required with a 4 : 1 FOS. Also, the 4 : 1 FOS was selected

for this example but does not necessarily mean that it is the standard. The designer

must check with local and in-house requirements for the proper FOS to use in his

situation.

Table 11.5 uses 80 % wire rope clip reduction for rope < 1′′ and reduction to 90%capacity for rope 1′′ and greater.

There are a few terms used with wire rope that need to be explained. The following

list describes these in no particular order.

IWRC: International Wire Rope Corporation sets the standards for wire rope

manufacturers.

XIPS: Extra improved plowed steel is the wire type used in the more widely used

wire rope in construction.

𝟔 × 𝟏𝟗: Wire rope designation gives the number of strands in a single wire rope

diameter and the number of individual wires within one strand. Therefore, 6 × 19

means there are 6 strands in the rope and each strand contains 19 wires. 6 × 19 is

a popular combination for construction use. Figure 11.20 shows the anatomy of a

6 × 19 wire rope’s cross section.

SWL: Safe working load is the allowable load of anything after an FOS has been

applied.


FIGURE 11.20 Wire rope anatomy.

Lay: A lay is the linear distance in inches along the length of the rope that it takes

for one strand tomake a complete 360∘ revolution. During themanufacturing process,

wire rope strands are twisted. This improves the strength, allows some stretch under

load, and helps keep the strands together.

Rule of Eighths for Sizing Wire Rope with a 6 : 1 FOS A “trick” or quick

calculation that was developed years ago in the field of construction enables one to

quickly determine the capacity of a certain diameter wire rope. The rule says:

1. Convert the wire rope size fraction to eighths if it is not already in eighths

(e.g.,1

4= 2

8,

1

2= 4

8, etc.).

2. Square the numerator.

3. Divide the new numerator by the denominator 8.

This is the wire rope capacity in tons with close to a 6 : 1 FOS. Table 11.6 shows

these values from1

4′′ diameter to 1′′ diameter.

TABLE 11.6 Rule of Eighths

Wire Rope Size Size in 8ths Square the Numerator SWL (tons)

1

4

2

8

4

80.5

3

8

3

8

9

81.1

1

2

4

8

16

82.0

5

8

5

8

25

83.1

3

4

6

8

36

84.5

7

8

7

8

49

86.1

18

8

64

88.0


TABLE 11.7 Wire Rope Termination Specifications

Wire Rope/

Clip Size

Minimum Number

of Clips

Turn Back

(in)

Torque

(ft-lb)

1

42 4.75 15

3

82 6.5 45

1

23 11.5 65

5

83 12 95

3

44 18 130

7

84 19 225

1 5 26 225

Source: Crosby Corporation, Wire Rope Clip Manufacturer.

This rule is very conservative and always produces a value that slightly exceeds the

6 : 1 FOS. Compare these SWL values to the first wire rope chart’s 6 : 1 FOS column.

The values should be close.

Properly Installing Wire Rope Clips The proper installation of wire rope clips

can be found in any manufacturer’s published data. Table 11.7 has been provided

here to explain the different specifications one needs to follow to make a proper wire

rope connection that will stay within 80% of the wire rope’s SWL capacity. For actual

specifications of wire rope accessories, the design engineer should refer to the product

data provided by his supplier.

The wire rope size usually, if not always, matches the clip size. The minimum

number of clips is how many clips are required to match the strength of the rope itself

and increases as the rope size increases. In order to fit the number of clips required,

the next column indicates the amount of rope “turn back” that is necessary so that the

clips are spaced properly and are not too close or too far from one another. Finally,

since each wire rope clip has two nuts that attach to one U bolt, when tightened these

nuts have a required torque value in order to match the capacity of the wire rope clip.

In addition to this information, the user should read the manufacturer’s installation

requirements, such as the proper orientation of the clip and the use of accessories like

thimbles (cable softeners).

Step 2 is complete. The next step is to determine the anchor block requirement.

Step 3: Determining the size and number of anchor blocks.

When starting this step, it should be pointed out that most companies already

have deadmen in their inventory that meet a particular size. For instance, one com-

pany interviewed for this chapter indicated that all their deadmen weighed 9600 lb.

At 150 pef concrete weight, this is the equivalent of 64 ft3 (4′ × 4′ × 4′ or equivalent).Figure 11.21 shows an example of how two deadmen were doubled up for twice the

sliding capacity. The wire rope was also run between the two so that tipping was

minimized.


FIGURE 11.21 Deadmen for a guying system.

This same 9600-lb deadman will be used throughout this step of the example. Theminimum weight of the anchor blocks will be determined, and then the quantity perwire rope connection will be known. This method is realistic and often uses a lotmore weight than required because the number of blocks is usually rounded up tothe nearest block. On occasion, the engineer will round down to the lower number ofdeadmen if it stays within a reasonable FOS. For instance, if an FOS of 2 was desiredand using only one deadman produced an FOS of 1.9, the engineer could possiblyaccept this condition, maybe knowing that there was redundancy somewhere else inthe design.

The forces that were determined earlier are going to be used in the design of theanchor blocks. For the required weight, the horizontal and vertical forces will be usedrather than the axial cable force. In addition, the coefficient of friction is required.This example will use a value of 0.40 for 𝜇. In Example 11.1, it was learned thatthe deadman weight was greatest when the pipe brace was pulling and lifting. There-fore, it would be an additional and unnecessary exercise to calculate the requiredweight while pushing the deadman. More importantly, since this example is usingcables, it would also be impossible to push the anchor considering cable cannot act incompression.With that said, using the FH and FV forces for each level anchor system,determine the minimum number of deadmen required.

Upper Level Deadman ΣFx = 0 and first solve for the normal force (N). FH ×FOS − ¯N = 0

(+3146 lb × 1.5) − 0.4N = 0 N = 11,798 lb

The forces can now be totaled in the y direction with a value for N.

11.4 ALTERNATE ANCHOR METHOD 289

ΣFy = 0, using the new value for N. Remember, N has the FOS built in so do not

apply another FOS.

+FVA −W + N = 0.

+3146 lb −W + 11,798 lb = 0 W = 14,944 lb

14,944 lb∕9600 lb∕deadman = 1.6 deadmen, use 2 ea − 9600 lb deadmen

Lower Level Deadman

(+5354 lb × 1.5) − 0.4N = 0 N = 20,078 lb

ΣFy = 0, using the new value for N.

+5354 lb −W + 20,078 lb = 0 W = 25,432 lb

25,432 lb∕9600 lb∕deadman = 2.6 deadmen, use 3 ea − 9600 lb deadmen

If two deadmen were used for the lower level guy cables, the FOS would decrease.

In order to know by how much, one would have to recalculate the required weight

without an FOS and then divide into 9600 lb × 2 = 19,200 lb and this would be the

new FOS.

(+5,354 lb) − 0.4N = 0 N = 13,385 lb

+ 5,354 lb −W + 13.385 lb = 0 W = 18,739 lb

FOS = 19,200 lb∕18,739 lb = 1.02, this is almost the equivalent of not having an

FOS at all, so using three deadmen in this case is necessary.

11.4 ALTERNATE ANCHOR METHOD

When the solution is not acceptable, as in the case of the number of deadmen needed,

as always in temporary structure design, the designer could go back into the problem

and rearrange some assumptions. Such as in the previous example, the guy cables

could be relocated vertically and the angle could even be changed, in hopes of low-

ering the forces and maybe using only two deadmen per guy level.

Another method of anchoring guy cables to the ground level could be accom-plished by using driven or drilled pile encased in concrete. Figure 11.22 is an

example of a set of guying cables attached to a cast-in-drilled-hole (CIDH) anchor.

Rebar hooks have been placed in the top of the anchor in order to have a place

for the screw pin shackles to connect. As a student of temporary structure design,

one should consider what would go into this design. What determines this pile’s

capacity?

Length of pile

Pile diameter

Skin friction between pile and soil


FIGURE 11.22 Drilled pile anchor.

Rebar size

Rebar embedment into CIDH

Strength of concrete

Angle of wire rope guying

Number of guys per CIDH

Most, if not all, of these make up the anchor capacity.

Anchoring guy wires to a concrete-encased pile has some advantages and disad-

vantages. The advantages are:

They take up less room.

They can be designed with more capacity (compared to more than two deadmen.

The disadvantages are:

They are more costly to install (an estimated $2500–$6000).

They also may have to be removed.

If they are not removed, the beams are liquidated.

Another item that bracing can be attached to is the shoring used for a coffer-

dam surrounding the foundation excavation if used. This could include sheet pile and

soldier pile. The design would have to be checked so that these piles can support the

bracing load as well as act as the support of excavation at the same time. A combined

stress analysis would also have to be considered.


Anchoring to Concrete Many times, the vertical elements being anchored or

guyed are part of a larger foundation slab on grade such as in commercial and treat-

ment plant work. In this case the bracing or guying is attached to a foundation using

plates, bolts, and nuts. It is assumed that the foundation is more than capable of sup-

porting the forces and the weak link is the connection hardware. This connection

could also represent how the pipe or cable is attached to the top of a concrete deadman

as well. Figure 11.23 would be an example of the two main forces being considered

at this connection point. Therefore, this calculation should be done in either case.

Table 11.8 has been included for drill-in anchor bolts in concrete, taking the aver-

age capacities of similar anchors available in the market. It is very important to know

the f ′c strength of the concrete at the time of anchor use. If the concrete strength has to

be estimated, use the more conservative value. Additional concrete cylinders can be

cast if the concrete strength during anchor installation and use is critical to the design.

The engineer of record would want to know this value before loading the system to

make sure its value is within the design criteria.

Since the drill-in anchor capacities are shown as “ultimate values,” the designer

must apply an FOS to the values shown. In the case of these anchors, a 2.0 : 1.0

FOS is sufficient. With this information, and the forces on the top of the deadman

FIGURE 11.23 Anchoring system to concrete.

TABLE 11.8 Tension and Shear Values for Drilled-in Concrete Anchorsa

Ultimate Tension and Shear Values for Drill-in Anchor Dolts (lb) in Concrete

F′c = 2000 psi F′

c = 4000 psiAnchor

Dia. (in)

Required

Torque (ft-lb)

Embedment

depth (in) Tension Shear Tension Shear

1

48 2

1

82,260 1,680 3,300 1,680

3

825 4 4,800 4,000 5,940 4,140

1

255 6 5,340 7,240 9,640 7,240

5

890 7

1

27,060 9,600 15,020 11,900

3

4175 6

5

810,980 20,320 17,700 23,740

7

8250 8 14,660 20,880 20,940 28,800

1 300 91

218,700 28,680 26,540 37,940

aValues in this table are industry averages.


in Examples 11.1 and 11.2, the designer should be able to size the anchors that willsufficiently support the connection of the wire rope or pipe brace to the top of thedeadman.

For each example, determine the bolts required to attach either the pipe brace orwire rope to a concrete slab or deadman that consists of 4000 psi concrete.

Example 11.3 Pipe Brace

FV = 2035 lb × 2.0 FOS = 4070 lb (tension)

FH = 3053 lb × 2.0 FOS = 6106 lb (shear)

From Table 11.8, the proper anchor is selected that can support the shear and

tension force at the same time. The tension force would require only a 3

8′′ anchor

(3300 lb < 4070 lb < 5940 lb). For the shear, since it is governing, the 1

2′′ anchor

would be required (4140 lb < 6106 lb < 7420 lb).

Conclusion: Use 1

2′′ anchor governed by shear.

Example 11.4 Wire RopeThe wire rope vertical and horizontal forces were the same because the cables were

placed at a 45∘ angle.

Lower Brace

FV = FH = 5354 lb × 2.0 FOS = 10,708 lb (tension and shear)

Again, from Table 11.8, select the anchor that can support both the shear and

tension force from the cable load. The tension and shear force would both require

the same anchor in this case, which is a5

8′′ anchor.

Conclusion: Use5

8′′ anchor governed by shear and tension.

Upper Brace

FV = FH = 3146 lb × 2.0 FOS = 6292 lb (tension and shear)

Finally, from Table 11.8, select the anchor that can support both the shear and

tension force from the cable load. The tension and shear force would both require the

same anchor in this case, which is a 1

2′′ anchor.

Conclusion: Use 1

2′′ anchor governed by shear and tension.

In many cases, the engineer or contractor may elect to use 5

8′′ anchors everywhere,

so there is no confusion of where the1

2′′ is used and where the

5

8′′ is used.

Internal Bracing Internal bracing is used to resist external lateral forces with-out extending guying or bracing beyond the footprint of the support structure itself.Internal braces are common with falsework frames (bents) as shown in Chapters 10


FIGURE 11.24 Falsework bent with 2% lateral force.

and this chapter. The bracing can be 2× wood members nailed or bolted to timberposts or rebar welded to steel pipe. Regardless of the type, the bracing should bedesigned to resist the external forces applied to the system. The external forces cancome from:

Wind

Earthquake

Impact from construction equipment or third-party (public) vehicular traffic

Wind on falsework would typically be low due to the small amount of surfacearea being contacted by the wind. However, sometimes a conservative approach istaken that applies the wind force to the side of the falsework as if it were a solid massfrom the ground to the top deck of the bridges. This was discussed in Chapter 10.Figure 11.24 illustrates a falsework bent with transverse bracing.

Earthquake forces have become better understood in the last several decades andseem to dominate (at least on the West Coast) the cause of lateral loads to our tempo-rary, as well as permanent, structures. A common force calculation takes the dead loadof the structure within the tributary area of the support being analyzed and induces2% as a lateral force at the upper portion of the structure. The sketch in Figure 11.24shows how this force would be applied to a single-support system.

Impact forces from equipment or traffic can be estimated with a few assumptionsand physics. However, this may not govern the worst-case scenario for the design.Also, other precautions are usually taken by the engineer and the contractor to avoidallowing this to control the lateral bracing. For instance, on state highway work, addi-tional struts are added to falsework bents, and loads/forces are doubled when thesesupports are near traffic. Sometimes this can double the number of falsework postsrequired.

Sample Calculation Bridge superstructure weight, 10,000 plf of bridgeWeight associated with the heaviest loaded support (34′), 10,000 lb/34 ft span =

340,000 lb2% of this weight as a lateral force, 340,000 lb × 2% = 6800 lbIf this 6800-lb force is applied to the top of the frame shown in Figure 11.23, how

big would the cross bracing need to be?Apply the 6800-lb force to either side of the frame. Then determine the slope of

the diagonal braces. In this case, the space between the posts is 16 ft and the distancebetween the top and bottom connections is 14 ft. Figure 11.25 illustrates this lateralforce and the dimensions of the bent.


FIGURE 11.25 Typical bent with internal bracing.

A force triangle can be created with the geometry of the height and width of thebracing from connection to connection. The 14′H × 16′W slope is used to developFV , FH and, ultimately, F (diagonal) force. By proportioning the known force(FH of 6800 lb) with the known force diagram dimensions, the FV and F forces canbe determined.

6800 lb

FV= 16′

14′

FV = 5950

F =√(6800)2 + (5950)2 = 9036 lb

Figure 11.26 shows the force diagram for this example as well as the vertical,horizontal, and diagonal dimensions.

The 9036-lb force is in tension as shown with the forces acting in the directionshown. Seismic forces can work in any horizontal direction in the way a wind loadcould act. If the 6800-lb horizontal force is reserved and acts from right to left,the vertical and diagonal force will be the same because the force triangle has notchanged. The only difference is that the opposite diagonal is in compression.

Wire Rope System If wire rope was used to brace this system, the wire rope wouldhave to be selected to match the 9038-lb force with a factor of safety of 2 or 3 : 1.Using Table 11.6 with a 3 : 1 FOS and 80% reduction for wire rope clips, the wirerope size can be determined:

9036 lb∕2000 lb∕ton = 4.51 tons

Use5

8′′ wire rope.

FIGURE 11.26 Force diagram and dimensions.


FIGURE 11.27 Guying frame.

Internal bracing can take the form of a four-sided frame, as in Figure 11.27, which

was taken on a project over a northern California river where there is no place to attach

the guy externally and the ground was too far below the work surface. The frame was

designed so that the rebar cages and formwork could be guyed to the frame, which

would transfer these forces through its components to something more substantial

below, such as a work trestle.

To be most economical, the frame should be designed to support all scenarios of

guying throughout the sequence of construction of one pier and, preferably, the reuse

on other piers that may differ slightly in geometry but require the same support. In

other words, it should be designed for as many reuses as feasible.

Longitudinal Bracing Longitudinal bracing connects one falsework bent to an

adjacent falsework bent in the longitudinal direction of the bridge or structure. These

braces should withstand the same 2% lateral force as the transverse bracing andwould

usually require that the stringers also be connected to the top cap so the forces are

transferred properly. Welding is always acceptable, but the preferred method would

be 2000-lb capacity beam clamps that attach the bottom flange of the stringer to the

top flange of the cap beam.

Wire rope attachments to top and bottom cap beams would usually come in the

form of shackles through a hole in the beam flange, wire rope, and wire rope clips

and with beam clamps at the upper stringer locations as shown in Figures 11.28

and 11.29.


FIGURE 11.28 Longitudinal bracing connected to bottom cap.

FIGURE 11.29 Beam clamp attaching a stringer to a top cap.


CASE STUDY: REBAR BRACING FOR BRIDGE PIER

Project Overview

Highway 162 in Oroville, California, had an old bridge section spanning the

Feather River that was built in the 1920s. In the latter part of the 20th century,

engineers were concerned that the center pier (in the middle of the Feather

River) was compromised and may not survive many more high-water experi-

ences. CalTrans decided to build a new bridge to replace the existing antiquated

bridge. In the fall of 1996 the new project was put out to bid and awarded. In

January of 1997, the Feather River came to within a few feet of cresting and

flooding Highway 70. The contractor and state agency grew more and more

concerned about potential risks in the next 3 years of construction.

As mentioned previously, the bridge piers were extremely large and awk-

ward. The base of the pier was 35 ft × 6.5 ft and the top (45 ft above water)

measured 70 ft × 6.5 ft. The pier reinforcing steel was prefabricated on site in

slightly less than half-sections. The hoisting of less than half the cage was done

by two cranes: Manitowoc 4000 and 3900W. The lift was designed so that the

larger crane would have the whole load eventually. The cage was very flexible

and internal bracing was added. This was all done by experienced iron workers

who had designed and built similar lifts on past projects. This was years before

state authorities began putting specifications together to reinforce cages such

as the one built for this project (Figure 11.CS1).

FIGURE 11.CS1 Hoisting of rebar cage.

(continued)


CASE STUDY: REBAR BRACING FOR BRIDGE PIER

(Continued)

Pipe braces were engineered and placed on the cage prior to hoisting into

place. These braces were positioned on the cage so that they could attach to the

top of the sheet pile cofferdam and be at an acceptable angle that would not

overstress the pipe braces. Since the braces were pipe, they acted as push-pull

bracing and could be placed on one side. The bottom of the cage rested on the

pier footing (Figure 11.CS2).

FIGURE 11.CS2 Installing braces.

FIGURE 11.CS3 Placing of intermediate hoop reinforcing.


The remaining rebar had to be placed by hand with crane assistance to attach

the two halves in the center. The forms were placed once the rebar was com-

pleted. About one-third of the pier’s top 70-ft dimension was tucked underneath

the existing bridge (Figure 11.CS3).

There were three piers in total. The same process was followed for all three.

The production improved slightly, but the methods did not change from the first

cage to the third cage.

CHAPTER 12

TRESTLES AND EQUIPMENTBRIDGES

Trestles or equipment bridges help contractors access their work in conditions that arenot conducive to standard methods. Trestles are used in construction to access workthat is not the typical flat land access, or it may be flat but environmentally sensitiveor over water. For these and other reasons, it is necessary for contractors to designtrestles and work bridges in order to perform the work. Access over water or roughterrain is fairly common in the heavy civil arena in order to get the resources close tothe work and reduce the need for floating equipment, such as derricks and barges, oreven “high-line” construction methods. High-lines are wire rope and pulley systemsthat span canyons, valleys, gorges, and even steep terrain in order to suspend blockand tackle that hoists materials to these remote areas.

Locations, such as these, can greatly affect access at projects such as the HooverDam, where the topography dictates a method other than a work trestle. Figure 12.1shows a Manitowoc 3900W working from a trestle on the Feather River in northernCalifornia.

12.1 BASIC COMPOSITION OF A STANDARD

TRESTLE

A trestle or work platform can be designed in all shapes and sizes. However, in theauthor’s experience, trestles fall within the same design format the majority of thecases. This format consists of driven pile, framework from wide-flange beams, andsmall miscellaneous metal sections and wood decking. This format will be describedin the following section, and the examples of this chapter will, for the most part,maintain this philosophy.

300

12.1 BASIC COMPOSITION OF A STANDARD TRESTLE 301

FIGURE 12.1 Driving sheet pile from work trestle.

12.1.1 Foundation—Pipe, H Pile, and Wide-Flange

and Composite Piles

Trestle foundations are rarely, if ever, founded on spread footings. More often, a tres-tle is supporting the weight of a 150,000- to 300,000-lb crane and requires substantialsupport, usually driven pile. Also, when over water, a trestle has to be on piles inorder to elevate above the high-water mark. So for several reasons, most trestles willbe always founded on driven or drilled pile. Finally, if we consider that these piles aretypically removed after the trestle life is over, then one would conclude that drivenpile (and usually steel pipe or beam) is more easily removed.

Piles for trestles have to withstand several forces, some of which are magnifieddepending on how high the trestle is above grade or dredge line. Typical forcesinclude:

Bearing and Friction: The pure dead load of the anticipated loads, which is notonly the crane itself but the crane’s weight while lifting its maximum load

Bending/Lateral Stability: Due to the dynamics of the crane or other equipmentmoving, starting, and stopping

Bending/Lateral Stability: Due to water movement, which could come from waveaction or flowing water such as in a river

Bending/Lateral Stability: Due to impact by marine equipment or vessels

12.1.2 Cap Beams—Wide-Flange Beams with Stiffeners

Cap beams are the lateral members above the pile that support the stringers or gird-ers. The cap beams transfer the stringer/girder loads into the pile. Cap beams can be

302 TRESTLES AND EQUIPMENT BRIDGES

FIGURE 12.2 Cap beam loading conditions.

designed so that they only transfer shear and reaction forces, or they can be placed in

a bending condition as well. Figure 12.2 illustrates these two scenarios.

Web yielding, as we learned in Chapter 7, occurs when a large load is transferred

from one member to another, in which case the supporting member’s web is suscepti-

ble to buckling (yielding). The probability of this occurring is higher when the stringer

(arrow) lines up with the pile (supports). In this case, the load is so high and the web of

the beam is so tall and/or thin that it buckles like a column. Stiffeners or gusset plates

are used to combat web yielding at high shear concentrations. High shear concentra-

tions occur at reaction points such as pile to cap beam location or stringer/girder to

cap beams. Typically, it does not take a very large gusset to resist the web yielding; it

just takes something added to the web to act like a small column, such as 1

2′′–3

4′′ thick

flat plate welded against the beams web between the top and bottom flanges. In the

case of falsework, contractors sometimes use small pieces of 4 × 4’s and wedge them

between the top and bottom flanges of the beam. Figure 12.3 shows a large stringer

supported by a stout cap beam. Figure 12.4 is a sketch of how web yielding would

affect a beam.

FIGURE 12.3 Stringer supported by cap beam.


FIGURE 12.4 Web yielding.

12.1.3 Stringers/Girders—Wide-Flange Beams Braced Together

Stringers or girders span from one bent cap to the next bent cap similar to stringers forbridge falsework. One difference between the stringers for trestles and stringers forfalsework is the lateral bracing between the stringers. Typically, stringers and girdersare designed for bending stress requirements and are then checked to see how muchbracing is needed to resist lateral buckling (which will be discussed next). Since thereare so many more dynamics and chances of impacts with trestles, the lateral bracingtends to be steel angle or channel welded to the inside web or flange of the beams.With falsework, as shown in Chapter 10, lateral bracing can be achieved using only4 × 4’s and steel banding since there is less chances of dynamics.

Depending on the depth of the stringers, they may also require stiffeners directlyover the cap beams similar to the stiffeners resisting web yielding. Stringers are oftenput together as “teams” prior to placement. Teams are two beams attached togetherby their lateral bracing (which is already required) so that they can be placed as aunit, thus requiring less handling of pieces at the job site. Teams are usually puttogether either off-site or on-site in a fabrication yard to maximize the efficiency ofthe welders.

12.1.4 Lateral Bracing

Once a beam is sized for bending stress, it should always be checked to see if bracingis required. This problem can be solved in two ways:

1. Compute the critical length (Lb) that the beam can span without bracing, andplace braces so that this length is not exceeded.

2. Compute the allowable stress that a beam can endure while complying withbending and lateral bracing requirements, and size the beam for both.

The second method will always result in a larger beam, but with no, or little, brac-ing required. The first method is usually more common because the difference in


TABLE 12.1 Lateral Bracing Cost Comparisona

Cost Comparison between Larger Beams and Bracing Smaller Beams

Larger Stringers without Bracing Smaller Stringers with Braces at1

3Points

There are five stringers required

They are W36 × 230 without braces

They are 40 ft long

Steel weight = 230 × 40 × 5 ea = 46,000 lb

At a steel price of $0.60 per lb = $27,600

There are five stringers required

They are W30 × 173 but they require 4 braces

They are 40 ft long

Steel weight = 173 × 40 × 5 = 34,600 lb

At a steel price of $0.60 per lb = $20,760

Add 4 braces at $500 ea (labor and material)

Total cost = $22,760

aThe extra steel required to avoid bracing cost the project an additional $4840. The cost of the extra steel

would become more economical if 14 or more braces were needed.

cost between adding some bracing or increasing steel weight so lateral bracing is

unnecessary is usually very significant. Table 12.1 compares these options.

Lateral Bracing Calculations to Avoid Flange Buckling According to

AISC, 9th edition formula (F1-8), the following length terminology is used:

Lc is the maximum unbraced length for 0.66FyLu is the actual unbraced length for 0.6FyLb is the actual unbraced length

Lb =20,000

Fy(d∕Af )

where Lb = critical length of beam span at which the beam is subject to lateral

buckling

Fy = yield stress of steel, i.e., A36 or 36 ksi

d = depth of beam from the beam charts in Appendix 1

Af = area of flange from the beam charts in Appendix 1

20,000 ksi = constant

In this instance, Af is the area of the compression flange in the chosen beam. The

larger this area, the more resistance the beam has to flange buckling. Since Af isdivided into d (depth of beam), the d∕Af value from the beam charts indicates what Luvalue to expect. As the beams get larger, the d∕Af becomes smaller, thus increasing

the Lb in the formula Lb = 20,000∕Fy (d∕Af ). The units of d∕Af are a value over

inches (in/in2).

If Lu < Lc,Fall = 0.60Fy = 0.60(36 ksi) = 21.6 ksi beam span is fine for bend-

ing only

If Lu > Lb, Fall = 12,000 ksi∕Lu(d∕Af ) = new allowable stress in ksi, which

is less than (0.60Fy), where Lu = is the actual span of beam and 12,000 ksi is a

constant.


Example 12.1 Lateral BracingAn A36, W24 × 146 spans 40 ft. Determine whether this beam needs lateral bracing;

and, if so, determine the lightest beam that would replace this beam and would not

need lateral bracing.

Step 1: Calculate the critical length of the beam and compare it to 40 ft.

Lb = 20,000 ksi∕[(36 ksi)(1.76)] = 315.7 in∕12′′∕ft = 26.3 ft <

40 ft. This beam needs one brace.

Step 2: Determine how many braces would be needed.

To determine how many braces are needed, take Lb and determine if

it is greater or less than 1

2the beam span. If it is greater, then one brace

is required with this beam. If not greater, then more than one is required.

If Lb is less than1

2the span, but greater than

1

3span, then two braces

are required. If Lb is less than1

3span, but greater than 1

4span, than three

braces are required. Table 12.2 compares Lb with L.

TABLE 12.2 Critical Length vs. Actual Length

Critical Length Braces

Lb > L No braces1

2L < Lb < L One brace

1

3L < Lb <

1

2L Two braces

1

4L < Lb <

1

3L Three braces

Lb <1

4L Change beam size

Figure 12.5 illustrates how these beams would be braced.

FIGURE 12.5 Cross bracing between stringers.

This W24 × 146 needs one brace because 26.3 ft is less than 40 ft but

greater than 20 ft (1

2L).

Step 3: If bracing is not desired, determine the lightest beam that works for bend-

ing and flange buckling by calculating a new Fall. For bending only, Fall

is 21,600 psi (or 0.60Fy). In order to omit bracing, the new Fall must be

used to determine the maximum allowable moment.


Fall = 12,000 ksi∕[(40′ × 12′′∕ft)(1.76)] = 14.17 ksi. Use this allow-

able stress to determine the maximum bending moment for this beamwith

this span.

For Lu > Lc; Mmax = Sx × Fall (use Sx for the W24 × 146);Mmax =371 in3 × 14.17 ksi = 5256 in-k.

For Lu ≤ Lc; Mmax = Sx × Fall (use Sx for the W24 × 146);Mmax =371 in3 × 21.60 ksi = 8014 in-k.

So what does this mean? This means that if the Lu (40-ft span) valueis greater than Lc, then in order to use the W24 × 146 beam, a maximum

moment of 5256 in-k cannot be exceeded. If Lu is less than Lc, then no

bracing is needed and the W24 × 146 is fine as long as it does not exceed

a maximum moment of 8014 in-k. For this to occur, the span or the load

or a combination of both, would have to be reduced.

12.1.5 Decking—Timber or Precast Concrete Panels

The decking is the component of the trestle that the equipment makes contact withand transfers the equipment loads to the stringers. Because all the equipment andconstruction materials are over the top of the deck, means should be taken to avoidhaving materials, soil, lubricants, fuels, and the like from getting between the deckingcomponents. It is common for the contractor, when over water or traffic, to have toseal the gaps between the voids in the decking. Sometimes plastic sheeting or filterfabric is used to keep contaminants from falling below the trestle. In extremely envi-ronmentally sensitive areas, sealant between the deckingmembers could be necessaryto make leaks impossible to pass.

The most common material used for trestle decking is timber and usually 12 ×12 × 30 ft timbers are combined into five-timber mats. The alternative would be toindividually place and remove single 12 × 12’s, which would cost extra in labor andequipment and cause excessive damage to the timbers. When used as combined mats(also known as cranemats), one riggingmoves five timbers at once, so they are movedaround more efficiently. The timbers are bolted together with through-rods and havepicking points so they can bemoved with a crane and not just a forklift. While movingmats with a forklift is okay, more damage occurs from the forks and reduces the lifeof the mats.

Another option, though not as common, is to deck the trestle with precast concreteslabs as shown in Figure 12.6. While this option may be more costly in up-front costs,if slabs are taken care of so they can be reused a number of times, it may become afeasible option. As inmost temporary structure decisions, themost economical optionusually is the best.

Impact Factor This text uses an impact factor multiplier to combat the unknownsof equipment dynamics when starting and stopping quickly. Traditionally, insteadof trying to calculate the actual forces created by these unfortunate events, the deadload is increased by at least 30% to account for additional forces. This multiplier iscalculated as follows.

Impact factor, IF = 50

125 + L

where L is the length of the span and 50 and 125 are constants.


FIGURE 12.6 Precast concrete trestle deck.

The impact factor is dependent on the span of the beam involved: therefore, the

longer the beams, the lower the factor. The impact factor used is never greater than

30%: therefore, if the IF calculates to anything greater than 0.30, then the impact

factor would be 1.30 minimum. However, if the IF is less than 0.30, then the actual

number would be used. The following examples should give the reader a good under-

standing.

Example 12.2 When a beam span is 35 ft long, the IF = 50∕125 + 35 = 0.3125,

use 1.30 as the multiplier.

Example 12.3 When a beam span is 50 ft long, the IF = 50∕125 + 50 = 0.2857,

use 1.286 as the multiplier.

The multiplier is used to increase the shear and bending stresses on the beams

in question. This is an additional factor of safety, but for impact and dynamics. The

standard FOS (e.g., Fb = 0.60Fy) will still apply to the steel and wood members. This

is not considered redundancy, these are two different factors.

Some equipment bridges that will support only slow-moving equipment, such as

crawler cranes, may not have an impact factor, or may have a slight impact factor such

as 5–10%. This is discussed among engineers quite often. Redundancy in design is

common and the question of whether to include impact factors or how much of an

impact factor to include will continue to be discussed. Most often, the engineer of

record and the check engineer come up with an acceptable solution that pleases both.


In Example 12.4, a true impact factor will be used for fast-moving vehicles. In the

other examples, 5–10% will be used to show the variances.

12.1.6 Environmental Concerns

When access is difficult on a project, it is often because the project site is on or adja-

cent to a body of water, wetlands, steep terrain, and the like. Such areas are commonly

environmentally sensitive areas as well. They may contain plant, fish, and wildlife

species; noise and/or vibration restrictions; or any of a number of other conditions.

In the Carquinez Straights during the Benicia Bridge construction in the early part

of this century, the construction operations included pile driving of cofferdam sheet

piling and large-diameter permanent pile. The impact of these elements, whether per-

manent or temporary to the project, had an effect on the marine wildlife. The owner

and contractor had to mitigate any harm to the affected species. A bubble curtain

was employed to dampen the vibration and sound waves through the water that could

cause harm to the marine environment.

When piles are driven into a riverbed, as shown in Figure 12.7, it is common to

have silt from the river bottom contaminate the water. When these areas contain fish

habitat, the fish and game or water quality control permits typically require the control

of this silt release, which can be accomplished by using some sort of silt curtain that

contains the silt in one area, as well as mitigation for any silt contamination that does

occur. Some permits that may determine the means and methods of the contractor are:

DWR, Department of Water Resources

FEMA, Federal Emergency Management Agency

FIGURE 12.7 Work trestle over the Sacramento River.


WQCB, Water Quality Control Board

DFG, Department of Fish and Game

As mentioned earlier in the chapter, the most common trestle uses timber mats on

steel wide-flange stringers, over steel wide-flange cap beams, supported by pile. The

Example 12.4 will follow this design.

CASE STUDY: ANTLERS BRIDGE

Project Overview

TheAntlers Bridge project replaced a 72-year-old, 1328-ft-long steel deck truss

bridge with a 1970-ft-long reinforced concrete cast-in-place segmental bridge.

The bridge, which spans the Sacramento River arm of Shasta Lake in northern

California, is a vital part of the Bureau of Reclamation’s Central Valley Water

Project in California. It provides water for agriculture, power generation, flood

control, recreation, and environmental needs. The water can be 120 ft deep at

the project location, and it is common for the water depth to fluctuate 100 ft in

a single season. The many environmental, design, and construction challenges

on the project included:

• Major interstate route

• Deep water design and construction

• Hazardous waste consideration

• Difficult access

• Limited construction staging areas

• Storm water pollution prevention

• Variable water depth consideration

• Environmental construction windows

• Pier height

• Challenging bridge demolition demands

• Weather extremes: high rainfall, freeze, and snow area

Access to Work

One of the more difficult and expensive challenges to overcome was lake

access. In general, the seasonal drawdown of the lake is predictable. The

Bureau of Reclamation is required to draw the lake down to an elevation 46′

below full pool every year due to flood control requirements. Agricultural

drawdown is also predictable. It is governed by irrigation and water supply

contracts between the Bureau of Reclamation and its clients, and generally

results in a drawdown of approximately 40′–50′. Depending on the lake level

at the beginning of the season, one can predict that the lake will be, at a

minimum, 50′ lower at the end of the season. The largest unknown is lake level

(continued)


CASE STUDY: ANTLERS BRIDGE (Continued)

recharge because the weather in the area varies considerably. In reviewing

historical water cycles, the low-water period can be as long as 2 months in wet

years and 5 months in dry years (Figure 12.CS1).

FIGURE 12.CS1 Trestle on north side of lake.

The facts pointed to many options, each having a certain degree of risk.

Option 1

If lake levels were to stay low for two consecutive years, the trestle could be

built 46′ lower. With a good deal of certainty, the trestle would be exposed for

5–6 months, and with an aggressive schedule the foundations for the main piers

could be completed.

A crane-mounted barge was going to be necessary on the project anyway

for delivery of material from one side of the lake to the other. The barge crane

could be used during the high-water periods (Figure.12.CS2).

Option 2

Build trestle to full pool elevation and guarantee access all year.

The contractor chose option 1. The water levels in the previous 2 years were

very lowwith the highest elevation not even reaching the flood control elevation

of 46′ below full pool. Figure 12.CS3 shows the lake level rising in the first


FIGURE 12.CS2 Trestle under construction—year 1.

FIGURE 12.CS3 Lake level rising.

construction season. If water levels were going to continue to stay this low,

12-month access was available. Another reason is the cost of constructing the

trestle to full pool elevation was more than double due to additional material

and bracing requirements.

(continued)


CASE STUDY: ANTLERS BRIDGE (Continued)

Unfortunately, the weather did not cooperate. Within a month of starting

trestle construction (December 2009), the weather changed and the lake level

increased and inundated the work. Within 2 months the lake level increased to

above the design trestle deck height and work stopped. That year the lake level

increased 127′ to full pool and did not recede to flood control elevation until

September 2010. It again increased above the flood control level by December

2010, allowing only 3 months of access to the uncompleted trestle. As a result,

the contractor’s only option was to redesign and raise the trestle to full pool

elevation.

Example 12.4 Trestle Design ExampleDesign an access trestle for concrete and earth-moving trucks given the following

criteria:

Project Features: Two-span bridge with two abutments and one center bent. Use

two stringers, 11 ft apart and a double bent cap. For other criteria use, the drawings

provided in Figures 12.8–12.15.

The truck has two axles and single tires front and rear. The rear axle load is

50 k and the front axle load is 40 k. These loads are split in half for left and

right wheels. Therefore, front-left = 20 k, front-right = 20 k, rear-left = 25 k, and

rear–right = 25 k. See the sketch in Figure 12.8 for clarity.

Design Criteria

Use timber decking and all steel members below the decking. Use A36 steel with

Fb = 0.60Fy and an additional impact factor (IF). For decking, use Fb allowable =1800 psi and Fv allowable of 160 psi. Figure 12.8 includes additional information for

design.

FIGURE 12.8 Example 12.4 Truck bridge details.

Step 1: Position the truck in detail A in Figure 12.8 so that the wheel locations

of the heaviest wheels (50-k axle, 25 k each) generate the maximum shear force in


the timber decking. This location, by rule of thumb, is when one of the wheels is

12 in from the stringer below. Technically, the wheel can be moved to within 2–3

in and a higher shear value can be achieved; however, 12 in is used to simplify the

distances to 1 ft increments. Figure 12.9 shows the 12-in distance from the centerline

of the stringer to the centerline of the wheel. Sum the moments using the two 25-k

loads and determine the reactions, which are the maximum andminimum shear loads.

Since the decking is specified to be timber, the size of the timber can be determined

by shear and then later checked for bending. Unlike steel, timber is susceptible to

shear failure just as much as bending failure.

FIGURE 12.9 Wheel loading for maximum shear.

Shear:

ΣM about left side, Rright = [(25 k × 1′) + (25 k × 8′)]∕11′ = 20.5 k.

ΣFv = 0, −25 k–25 k + Rleft + 20.5 k = 0, Rleft = 29.5 k.

The maximum shear load from this arrangement is 29.5 k at the left reaction (sup-

port stringer). Draw a shear diagram similar to Figure 12.10 to visualize this.

FIGURE 12.10 Shear diagram.

Step 2: Calculate the impact factor for this 11-ft span.

Impact factor = 50∕125 + 11′ = 0.37 > 0.30; therefore use 1.30 multiplier.

Step 3: Determine the maximum shear in one timber and size the decking based

on shear. Now consider that, based on the size of one of these wheels, one wheel will

make contact with at least two and maybe three timbers at one time, the tire pressure

will also have an effect on this as well. Assume a wheel contacts three 12′′ timbers at

once, then calculate the “new” maximum shear on one single timber with the impact

factor. Figure 12.11 is a representation of this loading.

FIGURE 12.11 Wheel loading on timber.


Vmax = 29.5 k × 1.30∕(3 timbers) = 12.8 k per one single timber.

Since Fv = 1.5V∕A for rectangular sections, then Amin = 1.5Vmax∕160 psi,

Amin = 1.5 (12.8 k)∕0.160 ksi = 120 in2

As a square section, this would require aminimumof 10.95′′ × 10.95′′. The closesttimber that is larger than these dimensions is an S4S 12 × 12. Now check to see if this

12 × 12 can withstand the maximum bending moment. An S4S 12 × 12 measures

11.25′′ × 11.25′′. Figure 12.12 shows a trestle with wood decking and hand railing.

This is a very common trestle arrangement between the decking and the stringers.

FIGURE 12.12 Trestle with wood decking.

Step 4:Assuming a 12 × 12, check bending stress. The section modulus of an S4S

12 × 12 is 237 in3. The allowable bending stress for the timber decking was given at

1800 psi. Now, position the truck so that the wheels generate the maximum possible

bending moment. The following rule explains how this is done.

Rule: To generate the maximum bending moment in a beam with two forces either

equal or not, calculate the center of gravity between the two forces (wheels). Then

position the centerline of the beam span in between the heaviest force/wheel load(either one if they are equal) and the resultant (center of gravity) of the forces/wheels.

Using this rule, position the truck accordingly. The center of gravity between the

left and the right wheel is in the middle because both wheels are 25-k loads. Convert

the 25-k wheel load into a single timber force considering the impact factor and the

three timbers it makes contact with. Call these forces V1 and V2.

V1 = V2 =25 k × 1.30

3 timbers= 10.8 k


Now place the vehicle in a location on the 11-ft span so the centerline of the beam

is between the 10.8-k wheel on either side and the resultant location, which is 5.5 ft

from the left or 5.5 ft from the right as shown in Figure 12.13.

FIGURE 12.13 FBD of truck position.

Left support to left wheel = 3.75 ft

Left wheel to centerline of span = 1.75 ft

Centerline of span to resultant = 1.75 ft

Resultant to right wheel = 3.5 ft

Right wheel to right support = 0.25 ft

All these distances should add up to the 11-ft timber span. If they do, calculate

the maximum moment generated by this configuration. First, redraw the free-body

diagram without the unneeded dimensions and remove the resultant arrow and beam

centerline as in Figure 12.14.



ΣM = 0, Rright = [(10.8 k × 3.75′) + (10.8 k × 10.75′)]∕11′ = 14.2 k

ΣFv = 0, Rleft = −10.8 k − 10.8 k + 14.2 k = 7.4 k

Vmax = 14.2 k

Mmax = 7.4 k × 3.75 ft = 27.75 ft-k

Solve for the bending stress in the 12 × 12 timber using a moment of 27.75 ft-k

and a section modulus of 237 in3.

Fb = (27.75 ft-k × 12′′∕ft)∕237 in3 = 1.405 ksi or 1405 psi < 1800 psi (OK)

Step 5: Determine the front and rear axle forces on one stringer using impact

factors and with the truck in the worst possible case for the stringer in order to design

the main stringers spanning 80 ft, as shown in Figure 12.15. Since the beams are

simply supported single spans, the design of one will represent both of them. Start by

calculating a new impact factor for the 80-ft span.

IF = 50∕125 + 80 = 0.24

FIGURE 12.15 Truck positioned on 80-ft span.

Since this value is less than 0.30, use 1.24 as the multiplier. Now multiply this

value times the maximum wheel load on the girder in the worst-case scenario. This

worst-case scenario was determined in the timber design portion of this example,

which used a 29.5-k load in order to generate the highest shear. In other words, the

stringer experiences the heaviest load when the truck drifts to the left or right and

gets within one foot of the centerline of the stringer. Another thing to consider is that

if we use 29.5 k instead of 25 k, we should use a higher load that the other 40-k front

axle load. First, calculate the heavier load with the 1.24 impact multiplier, and then

figure out what the front axle load will be.

Rear = 29.5 k × 1.24 = 36.6 k

Now for the front axle single wheel load on the stringer, use a proportioning

technique from the rear to the front. We know that the original rear load for a sin-

gle wheel was 25 k and the front was 20 k. Proportion these figures to achieve the

missing front axle load, considering worst-case positioning of the vehicle and the

impact factor 36.6 k is to 25 k as (front axle force, Ff ) is to 20 k; front axle force =36.6 k × 20 k∕25 k = 29.3 k

36.6 k

25 k=

Ff20 k


Now both forces from either the left or right side of the truck are known to be on

one single stringer, using the worst-case position and the impact factor.

Step 6: Stringer design. Using the forces from step 5, size the stringer for bending

stress (ignoring shear) and then check for flange buckling. The rule that was used

previously will be used again. The rule has been restated below with customization

for the stringer design and two unequal loads, front and rear.

Rule: To generate the maximum bending moment in a beam with two unequal

forces, calculate the center of gravity between the two forces (front and rear). Then

position the centerline of the beam span in between the heaviest force/wheel load(rear in most cases) and the resultant (center of gravity) of the two forces/wheels.

Determining center of gravity for two unequal forces requires that the moments

be added in order to find the location of the center of gravity. The forces involved are

the rear 36.6-k force and the front 29.3-k force. The total resultant force from these

two are 36.6 + 29.3 = 65.9 k. The axle spacing of this truck was given at 24 ft wide.

Therefore, the center of gravity will be somewhere between the middle and the 36.6-k

load. Choose one side and add the moments, using the wheel loads and the resultant

force of 65.9 k.

ΣM (rear wheel) = 0

(65.9 k × X′) = (29.3 k × 24′), where X′ is the distance from the rear wheel and the

resultant:

X′ = 29.3 k × 24′

65.9 k= 10.7 ft

Figure 12.16 indicates the dimensions between the resultant and the front and rear

axles.

FIGURE 12.16 Resultant location.

Now follow the rule for maximummoment placement and set the centerline of the

80-ft-span between the 36.6-k load and the resultant. Divide the 10.7 dimension in

half and place the centerline 5.35 ft from the 36.6-k load.

Redraw the free-body diagram with only the information needed and make sure

all three dimensions add up to 80 ft. See Figure 12.17 for the proper placement of the

truck on the span.

ΣM = 0, Rright = [(36.6 k × 34.7′) + (29.3 k × (34.7′ + 24′)]∕80′

= 37.4 k

ΣFv = 0, Rleft = −36.6 k − 29.3 k + 37.4 k = 28.5 k


FIGURE 12.17 Placement of truck on 80-ft span.

Vmax = 37.4 k

Mmax = 28.5 k × 34.7 ft = 989 ft-k

Using Fb = 21.6 ksi for allowable bending stress and 989 ft-k for maximum

moment, determine the three lightest W shape beams that will work. Then check all

three for flange buckling and determine the bracing needed.

Sx(min) =(989 ft-k)(12′′∕ft)

21.6 ksi= 550 in3

Looking at Table 12.3, all three of these beams were selected on the basis of bend-

ing requirements. Two of the three need four braces according to Lc requirement, and

the other, five braces. The braces are recommended at each end and intermediates are

required per the Lc calculation. Based on its need for an additional brace, it might

seem logical to rule out the W30 × 191. The other two beams are very different in

weight. In this case, using the lighter of the two beams (W27) would seem to make

sense because it satisfies bending requirements and needs the same number of braces

to install. With that said, we could have considered the W30x191 as good as choice

2 (W24 × 229) because the difference in weight between the two at $0.60/lb could

easily pay for the extra brace. A simple analysis, using the cost of $1000 to furnish,

install, and remove a single brace, tells the whole story.

TABLE 12.3 Flange Buckling Check for BEST beam

Beam Sx (550 in3min) d∕Af Lc (ft) Braces, Conclusion

W27 × 194 556 1.49 28.3 80∕28 = 3, 4 braces

W30 × 191 598 1.72 24.5 80∕24 = 4, 5 braces

W24 × 229 588 1.15 36.6 80∕36 = 3, 4 braces


W27 × 194, 80 ft × 194 = 15,520 lb × $0.60 = $9312 plus four braces @

$1000∕ea = $13,312

W30 × 191, 80 ft × 191 = 15,280 lb × $0.60 = $9168 plus five braces @

$1000∕ea = $14,168

W24 × 229, 80 ft × 229 = 18,320 lb × $0.60 = $10,992 plus four braces @

$1000∕ea = $14,992

As suspected, based on this cost analysis, the W27 is indeed the most economical

choice, the W30 was second best, and the W24 was third best.

Step 7: Design the center bent. Determine the cap beam sizes and the pile load

requirement. Ignore the abutments, as it is assumed they are on mudsills directly

on the ground surface. The pile sets at each of the individual bent caps are spaced

at 5-ft centers. The force arrow is the stringer load bearing on the bent cap beam

(see Figure 12.18). The first question is: What is the largest load one bent cap will

support? This can be answered by going back to the stringer design and positioning

the stringer close to the cap beams.

FIGURE 12.18 Stringers loading cap beams.

It can be assumed that only one truck is on the bridge at any given time or it can

be assumed that two could mistakenly be on the bridge, one per span, at the same

time. The latter, more conservative approach is usually always taken and will be in

this example. The reactions from step 6 indicate that 37.4 k was the reaction for one

stringer, so if there are two stringers, then the total load on the bent cap is twice the

stringer reactions.

Bent cap load = 37.4 k × 2 = 74.8 k


The maximum moment on the beam then isMmax = PL∕4; M = (74.8)(5 ft)∕4 =94 ft-k.

Vmax = 74.8 k∕2 = 37.4 k

Using the same allowable bending stress of 21.6 ksi, the minimum section modu-

lus is

Sx =(94 ft-k)(12′′∕ft)

21.6 ksi= 52.2 in3

Select two HP shapes that have a minimum section modulus of 52.2 in3 and then

check for web yielding. The two H-Piles meeting this criteria are listed in Table 12.4.

TABLE 12.4 HP Shape Selection

H-Pile Sx (in3) k tw

HP10 × 57 58.8 1.0625 0.565

HP12 × 53 66.8 1.125 0.435

Step 8: Web yielding.

The potential yielding occurs when a concentrated load or reaction force occurs

and the beam’s web cannot support the load and, therefore, becomes overstressed.

The allowable web yielding stress for any beam according to AISC, 9th edition, is

Fwy (allowable) = 0.66Fy.

A36 steel, Fwy (allowable) = 0.66(36 ksi) = 23.76 ksi.

Grade 50 steel, Fwy (allowable) = 0.66(50 ksi) = 33 ksi.

There are two beams in Table 12.4 that are adequate to resist the bending moment.

The beams have been determined to have a high enough section modulus to withstand

any bending stress. However, is either beam susceptible to web yielding? The best

beam, because of its weight per linear foot, is the HP12 × 53, so this beam should be

checked for web yielding. If it passes the check, then it will be the best beam. If it

does not pass the web yielding test, then the HP10 × 57 should be checked. If neither

beam passes, the designer can (1) add stiffeners or (2) select a beam that has a high

Sx and better web yielding resistance, which usually means a thicker web.

Web yielding occurs at the area of the web thickness and the length developed

from the width of contact (beam flange or pile width) and five times the K dimension.

K is designated in the beam values (see Appendix 1) and is the distance in inches

from the top of a beam to the point of tangent where the steel has a radius to the

web thickness. The concentrated load is a force acting downward from a beam above

and projects the force at an angle of approximately 22∘ (2.5 : 1). The beam’s flange

width dictates where this angle of projection begins. For a good understanding of this

concept, see Figure 12.19.

Scenario 1: Downward Projection

Fwy =P

[(Fb + (5K)) × (tw)]≤ 0.66 ksi


FIGURE 12.19 Web yielding downward projection.

where P = concentrated force or reaction

fb = flange width of upper beam

tw = web thickness of lower beam

K = dimension from top of lower beam to the tangent of the web thickness

(see beam values in Appendix 1)

Fy = yield strength of steel

0.66 = allowable stress multiplier

Scenario 2: Upward Projection (refer to Figure 12.20)

FIGURE 12.20 Web yielding upward projection.

The stress is a normal case as in P∕A. The difference is that A = [ fb + 5(K)] × tw.This is the rectangular section of the yielding section at the web located distance Kfrom the top of beam.


When the concentrated load is a reaction and comes from underneath the beam in

question, project the force at a 2.5 : 1 angle upward. Instead of fb for the upper beam,

use the dimension of the plate and/or pile width.

Step 8 (continued): Now, using this same principle in scenario 2, calculate the

stress on the pile cap beam from the concentrated load at the pile.

P = 37.4 k

Cap beam = HP12 × 53

tw = 0.435 in

K = 1.125 in

pile = HP14 × 89

fb = 12.695 in

d = 13.83 in

Depending on the pile orientation, either dimension can represent the beginning

of the projection.

Assuming the pile is turned so the fb dimension is transverse to the cap beam, use

the fb = 12.695 in and no plate is used.

Fwy =37.4 k

[12.695′′ + (5 × 1.125′′)] × (0.435′′)= 4.69 ksi < 0.66Fy (or 23.76 ksi) (OK)

Example 12.5 Crane TrestleA crane trestle is needed for a Manitowoc 4000W over a river. The river is normally

8 ft deep (20 ft during flooding) and flows at a rate of 9 ft/s. The work can only be

done in the low flow time, so 8 ft is the most the trestle will experience. The crane

has to access a pier in the middle of the river approximately 150 ft from the normal

water level (spring, summer, and fall) shore line. In the past, your company has built

trestles for similar size cranes using spans between 28 and 34 ft long. At 30 ft, this

would create five spans of trestle. This will be the goal since the available beams

in the company’s yard average 40 ft in length. Longer spans could be checked, but

a four-span trestle would require 37.5-ft spans (150/4). This would not allow the

stringers to have proper overlap on the cap beams, so five spans is the best option.

The maximum load the crane will lift during the life of this trestle is 39.9 k. Some

other characteristics of this lift are as follows. The boom length required in order to

reach the work during all phases of construction is 140 ft. All cranes have charts that

indicate the maximum capacity of the crane in multiple cases of boom length, radius,

and angle. Crawler cranes have two cases of maximum capacity, with the tracks

extended or retracted. It is most common to operate cranes with the tracks extended,

unless there are tight access restrictions. In this case, the trestle will be constructed

to accommodate the crane with extended tracks and a walkway for workers. The

crane radius (distance between the crane rotation and the load during maximum lift)


with 140 ft of boom and lifting 39.9 k is approximately 65 ft. The track dimensions

are 20.5 ft long (center of roller to center of roller) and 4 ft wide. The crane weighs

330,000 lb.

The total load of the crane while lifting 39.9 k is 369,900 lb (rounded to 370,000 lb

or 370 k).

Operating over the Side

The crane will be operating over the side while servicing the pier construction. There-

fore, this design analysis will be with the 39.9-k load over the side at a 65-ft operat-

ing radius. When a crane is lifting a maximum lift at its maximum radius, 80–90%

(tipping) of the total load shifts to the side of the lift. In this case, the more conserva-

tive 90% will be used. A value as low as 80% can be used if the designer wants to be

aggressive.

Use 90% tipping:

90% of 370 k = 333 k, 333 k∕20.5 ft (track length) = 16.24 klf, 16.24 klf∕4 ft(track width) = 4.1 ksf.

Operating over the Front

The gradient (triangular loading diagram) for the maximum load over the front of

a crawler crane is developed by determining how much of the tracks overall length

(20.5 ft) receives the load. One rule of thumb is that one-half of the track’s length

receives 100% of the load as a varying uniform load (triangular). Figure 12.21 shows

what this loading diagram would look like along one-alf of the 20.50-ft-long track

length.

Ll = Lt∕2 = 20.50 ft

2= 10.25 ft

FIGURE 12.21 Crane track loading over the front.

Another method is to calculate the moment that occurs during the maximum lift

(39.9 k × 65 ft) and then to find the eccentricity of the moment and back into the

length of track receiving the load:

e = MP

where e = distance from the resultant of the triangle to the centerline of the

loading diagram

M = moment generated from the maximum lift

P = total load between the crane and the lift (370 k)


M = 39.9 k × 65 ft = 2594 ft-k

e = 2594 ft-k

370 k= 7.0 ft

Ll = 3

[(Lt2

)− e

]

where L2 = loaded track length

Lt = length of the track

L2 = 3[(

20.50 ft

2

)– 7.0 ft

]= 9.75 ft

The maximum load of the varying uniform diagram (W) is the total track load of

333 k (from above) divided by the loaded track length (L).Method 1:

W = P∕L1,330 k

10.25 ft= 32.2 klf

(32.3 klf

8 ft= 4.0 ksf

)Method 2:

W = P∕L2,330 k

9.75 ft= 33.8 klf

(33.8 klf

8 ft= 4.3 ksf

)The total klf was divided by 8.0 ft, which represents two track widths. Since it

was decided earlier to use method 1 (the simpler method), the gradient for the front

loading would be 4.0 ksf at the high end and 0 ksf at the low end over a 10.25-ft

length. This is illustrated in Figure 12.22.

FIGURE 12.22 Crane track loading method 1.

Since the side loading was going to be the determining factor, the rest of the prob-

lem will use 16.24 klf and 4.1 ksf, respectively. This was the side load value using

90% of total load.

Timber Decking

If timber decking is used, the decking should be checked for shear and bending

resistance. Maximum shear allowable stress (Fv) is 140 psi and maximum bending

allowable stress (Fb) is 150 psi. The crane should be positioned in order to generate

the maximum shear on the timber. As a rule, it was discussed earlier in this chapter to

place loads 12 in from the reaction point and test the shear at this point. The contractor


has a large number of 8 × 18 timbers available in the yard; therefore, these will be

the most desirable timbers to use. The timbers will be full-sawn, which means their

actual and nominal dimensions are the same. Check the shear on an 8 × 18.

8 × 18 properties (full sawn)

A = 96 in2

Sy = 128 in3

16.24 klf∕4 ft = 4.1 ksf × 18′′∕12′′∕ft = 6.2 k

fv = 1.5(6.2 k∕96 in2) = 0.0.96 ksi or 96 psi < 140 psi ( OK)

Demand = 96 psi∕140 psi = 0.69

Bending over a 5-ft span between stringers, the beam program is used to generate

the maximum moment (see Figure 12.23). Note that the section modulus listed for

the 8 × 18 timber is in the “weak” direction (Sy).

fb = M∕Sy, fb = (12.3 ksi × 12′′∕ft)∕128 in3

= 1.153 ksi (or 1153 psi) < 1500 psi (OK)

Demand = 1153∕1500 = 0.77

FIGURE 12.23 Shear and moment diagram of single track.

12.1.7 Stringer Design

We can assume one stringer takes the total track load or we can distribute a large

percentage, say 75%, of the load to one beam. The more conservative approach is

usually appreciated when dealing with human lives and equipment that costs hun-

dreds of thousands of dollars. Place 100% of the track load onto the 30-ft stringer

span and determine the maximum moment for this condition. The beam software has



generated a maximum moment of 1680 ft-k (see Figure 12.24). Add an impact factor

to this moment.

IF = 50∕125 + 30 = 0.32, use 1.30 maximum

1680 ft-k × 1.3 = 2184 ft-k

Stringer Size To size the stringer, use the maximum bending moment determined

by the shear and moment diagram. Stringer size depends on Sx,min = (2184 ft-k ×12′′∕ft)∕21.6 ksi = 1213 in3 minimum section modulus without considering the

compactness of the beam. The beam will be used in the strong direction. Table 12.5

lists potential stringers.

None of the beams listed require bracing. Therefore, the lightest of the three

choices would be preferable. Remember, this is a conservative approach because the

whole track load was placed over one beam. In reality, the second stringer would

pick at least 25% of the load, thus reducing the moment by this same amount. In this

case the stringer requirement could have been as low as 900–920 in3 section moduli.

This is a designer’s decision and should be looked at closely.

As far as the bracing, this does not mean that braces will not be used at all in order

to “team” the beams for placement purposes. As mentioned earlier, this is a technique

contractors use so they can set two beams at one time, transport them together (which

require fewer picks), weld in more favorable conditions, and avoid having to stabilize

them temporarily during placement. This method is discussed in the trestle case study

on the Golden State Bridge.

TABLE 12.5 Potential A36 Stringers

Beam Sy (in3) d∕Af (∕in) Lc (ft) Bracing (ea)

W30 × 391 1250 0.87 48 None

W33 × 354 1230 1.06 39.7 None

W36 × 328 1210 (close) 1.21 34.8 None


Example 12.6 Truck or Rough Terrain (RT) Hydraulic Crane TrestleDesign the decking and the stringers for the information given below. A plan view

and side view have been provided in Figure 12.25.

FIGURE 12.25 Plan and side view of hydraulic crane.

A truck or RT crane gets its support from outriggers. Outriggers are hydraulic

extensions that widen the footprint of the cranes base and lift the crane from the

ground so that only the outriggers make contact with the ground. At this point, the

tipping potential for the crane has been reduced because the base dimensions have

increased. Outriggers are typically founded on round or square distribution plates that

spread the concentrated load over a 1- to 2-ft dimension. It is very common and almost

mandatory that these outrigger pads set onwood cribbing in order towiden the contact

area even more so that the supporting surface (ground or trestle) does not experience

high point load pressure (P∕A). Decent ground conditions can support from 2000 to

4000 psf pressure on average, and trestle decking can handle 140–160 psi in shear,

500–650 psi in perpendicular crushing, and 1500–1800 psi in bending.

Given the following information:

Weight of crane (without load) = 250,000 lb (50 k counterweight, 200 k

carrier weight)Max pick = 80,000 lb

Max radius with max pick = 40 ft

Boom works, rigging and wire rope = 20,000 lb@ 20-ft

radius center of gravity

Stringers are spaced at 5 ft OC and one outrigger (4 × 4) is supported in between

the two.

12 × 12 wood decking (Fv = 140 psi, Fb = 1500 psi)A36 steel, Fb = 21.6 ksi, Fv = 14.4 ksi

Outrigger pad = 4′ × 4′

Space stringers 5 ft apart and span 34 ft

To start, the reactions under each outrigger need to be determined at the point when

the heaviest load is being hoisted.

Add the moments about a given point. In this case, the rear outrigger location will

be fine.


(200 k × 2 ft) + (20 k × 20 ft) + (80 k × 50 ft)∕20 ft (outrigger spread) = 240 k

on front outrigger (FOR)Add the vertical forces.

Rear outrigger (ROR) = +240 k – 50 k – 200 k – 20 k – 80 k = 110 k

Since there are two outriggers in the front and two in the rear, divide these reactions

in half for one outrigger.

Front outriggers (FOR) = 240 k∕2 = 120 k (Figure 12.25)

ROR = 110 k∕2 = 55 k (Figure 12.26)

FIGURE 12.26 Diagram of crane showing outriggers.

The heaviest outrigger load (120 k) should be placed on the wood decking so the

decking material must support the outrigger load in between the stringers. The FBD

would look like Figure 12.27 with the uniform load being 120 k∕4 ft wide pad =30 klf, the width of the pad = 4 ft and the stringer spacing = 5 ft. See Figure 12.27

for the loading, shear and moment diagram.

FIGURE 12.27 Loading diagram of outrigger.

In order to check the decking material for stress in shear and bending, calculate

the maximum shear and bending moment, then check the stresses compared to wood

decking that is good for Fv = 140 psi and Fb = 1500 psi as given above.

Shear and moment diagram:

Maximum shear = (30 klf × 4 ft)∕2 = 60 k

Maximum moment = area under the shear diagram from the left support to the

center of the outrigger.

Mmax = (60 k × 0.5 ft) + (60 k × 2 ft∕2) = 90 ft-k


Check shear over 4 ft of timber (use 11.5′′ as actual dimension)

Cross-section area of a 12 × 12 = 11.5′′ × 11.5′′ = 132.3 in2 × 4 = 529.2 in2

fv = 1.5 (60 k∕(4)132.3 in2) = 0.170 ksi or 170 psi > 140 psi

(demand = 170∕140 = 1.21) (No Good)

Check bending over 4 ft of timber (use 11.5′′ as actual dimension)

Section modulus of the same 12 × 12 = (11.5′′)(11.5′′)2∕6 = 253.5 in3 × 4 =1014 in3

fb = (90 ft-k)(12)∕1014 in3 = 1065 psi < 1500 psi (OK)

The calculations indicate that the wood decking is failing in shear (170 psi), even

though it is fine in terms of bending. Therefore, a change has to be made in order for

both checks to work. The options are:

Use a larger timber. This option is difficult because the most common (and eco-

nomical) timber is a 12 × 12.

Reduce the spacing of the stringer. This was set as a given piece of information.

If the stringer spacing changes, this will change the stringer design. This is a potential

option.

Widen the outrigger pad to extend over the 5-ft beam spacing using additional

wood cribbing. This option is least expensive and allows the beams to remain at a

5-ft spacing. The amount of cross-sectional area of timber needed to lower the shear

stress has to be enough to lower the stress by 30 psi.

A = 1.5 (60 k)∕0 .140 ksi = 643 in2 − (4 × 132.3 in2) = 114 in2. If we divide this

by the 4-ft (48-in) outrigger width, then we can back into the thickness of 2.39 in.

This is the same thickness and same type of wood that would have to be added to

the decking in this location. A 4× pad, the same width of the outrigger, would be

adequate since it measures 3.5 in in thickness as shown in Figure 12.28.

FIGURE 12.28 Outrigger on 4× wood cribbing.

The next step would be to position the crane on a set of stringers to generate the

maximum bending moment for the stringer. Some assumptions can be made, such

as the outrigger load (centered on two stringers) will be divided in half. Apply half

the load to one stringer and half the load to the other. Draw a sketch of how the

crane is positioned on the trestle and the location of the stringers. Also, the maximum

outrigger loading on one set of stringers (either left or right) is when the crane swings

around and picks the load off the side of the trestle. Place the two 120-k loads on one

pair of stringers as shown in Figure 12.29.

The next question should be: Where do we place these two 120-k loads to generate

the largest bending moment on the stringer? Like Example 12.4, which had unequal


FIGURE 12.29 Plan view of crane outrigger pattern.

wheel loads, this will place the centerline of the stringer in between one outriggerand the resultant, which is the center of the outriggers. The outrigger spacing from

front to back is 20 ft.

A beam program was used to check the outcome of this crane positioning and

determined the Mmax moment to be 508.2 ft-k (Figure 12.30). Since the loads were

already divided by two for one individual stringer, this moment can be used to size

one single stringer.


Impact Factor

Previously, an impact factor was calculated for Example 12.4 involving a truck trestle

on which trucks stopping and starting hard is very likely. If we look at the same possi-

bility for a crawler or hydraulic crane, it should be obvious that the likelihood is much

less. Some engineers would not use an impact factor at all in this case. Others may

use 5 or 10%. Let’s calculate the impact factor the way we learned in Example 12.2

and 12.3 but use a much lower factor.

IF = 50∕125 + 34′ = 0.314. An IF of 1.3 would be the maximum. However, an

acceptable IF would be 10% for this condition because the truck and hydraulic cranes


are still going to be driving on their rubber tires while positioning.

Mmax = 508.2 ft-k × 1.10 = 559 ft-k

Sx,min = (559 ft-k × 12′′∕ft)∕21.6 ksi = 310.6 in3

Table 12.6 summarizes beam options with a flange buckling check.

TABLE 12.6 Beam Summary with A36 Steela

Beam Sx(in3) d∕Af Lc Bracing (Lu = 34′)

W18 × 175 344 1.11 455.5′′∕12 = 37 ft None

W21 × 147 329 1.53 330.4′′∕12 = 27 ft One in center

W𝟐𝟒 × 𝟏𝟑𝟏 329 1.98 𝟐𝟓𝟓.𝟑′′∕𝟏𝟐 = 𝟐𝟏 ft One in center

aBest beam is in boldface.

The best stringer appears to be the W24 × 131 without doing a cost analysis

between stringer weight and lateral bracing labor and materials. The beam is much

lighter than the W18 × 175 and requires the same number of braces as the heavier

W21 × 147 (one in center). Remember that this beam was designed based on the two

outside stringers taking the entire load. The other two outside stringers would be in

the same condition if the crane went to that side of the trestle to make a similar lift.

The center girder, however, is never going to see the same loads, other than some

wheel loading from trucks, forklifts, and the like. If that is the case, this stringer can

be much lighter than the four outside stringers.

Cap Beam and Pile Design

Position the crane to generate the largest end reaction on the stringers so that the cap

beam experiences the maximum load. At the same time, swing the crane 90∘ so that

the maximum load is being lifted over the front or rear, and not over the side as shown

in Figure 12.31.

FIGURE 12.31 Hydraulic crane positioned over cap beam.


The pile should be located under the cap beam at a spacing that minimizes the

bending moment in the cap beam and also maintains a manageable pile load. The

beam program will be used to find the best pile location in order to achieve this best

economy. It will be assumed that two piles are sufficient and that bracing can be added

if the piles are unstable due to water flow and seismic shifting at 2% dead load.

The beam program results that are most desirable are shown in Figure 12.32 with

an accompanying sketch. The first beam results represent the track loads (120 k/ea)

over the five stringers, both loads more concentrated over stringers 1, 2, 4, and 5.

The track loads are actually creating an “uplift” reaction at the center stringer (3) at

the time the crane’s heaviest outriggers are over the cap. The four outside stringers

are supporting the complete crane load. The piles are eventually going to carry the

complete 240-k crane load, 120 k per pile.

FIGURE 12.32 Outriggers on cap beam.

The beam shear and moment diagram above calculated the stringer loads. The

calculated loads from this diagram can be used on the second diagram (Figure 12.33),

which requires the stringer loads and the pile locations in order to check the cap beam

loading.

This beam shear and moment diagram shows the five stringer loads with the center

(3) acting upward. The piles are showing a force equal to the crane pad forces at 120 k.

The pile will require this force in axial load, while at the same time resisting bending

from water flow and seismic shifting. This combined stress analysis will be done in

the next step of pile design.

Pile Design

The total live and dead load on the pile cap should be accumulated in order to deter-

mine howmuch force acts laterally when considering a 2–5% lateral force on the total

weight over a single pile. The percent of dead load used varies between falsework


FIGURE 12.33 Cap beam on pile.

and equipment trestles in the industry. The minimum is 2%, and actually 3% can be

achieved from the dynamics of the equipment using published manufacturers’ equip-

ment data coming from speeds, start and stop forces, and swinging. For trestles, many

engineering firms use forces from 3 to 5% of dead load to take this into account, plus

add an additional small factor of safety. In this text, 2% will be used.

The weight is the total of the crane weight with the load it is carrying plus the

dead load of the timber, steel stringers, cap beam, and any other miscellaneous dead

load weight that the pile will have to support. Figure 12.34 shows this condition as a

combined stress load.

FIGURE 12.34 Pile combined loading.

Pile Load

Live crane load—all four outrigger loads = 350,000 lb

Live personnel load—28′ × 34′ × 50 psf = 47,600 lb

Stringers—5 × 38′ × 131 lb∕ft = 24,890 lb


Decking—28′ × 34′ × 35 pcf = 34,320 lb

Total live and dead load = 456,810 lb

Distributed to two pile (divide by 2) = 228,405 lb per pile

Lateral seismic load at 2% of DL + LL = 4568 lb (see Figure 12.35)

(see also Table 12.7)

FIGURE 12.35 Seismic force at top of pile.

TABLE 12.7 Common Pipe Used as Trestle Pile

Pipe Size

OD

(in)

ID

(in)

Area

(in2)

Section

Modulus (in3)

Moment of

Inertia (in4)

r (in)

18′′ × 0.25′′ 18 17.5 13.94 61.0 549.1 6.28

18′′ × 0.50′′ 18 17 27.49 117.0 1053.2 6.19

20′′ × 0.25′′ 20 19.5 15.51 75.6 756.4 6.98

20′′ × 0.375′′ 20 19.25 23.12 111.3 1113.5 6.94

24′′ × 0.25′′ 24 23.5 18.65 109.6 1315.3 8.40

24′′ × 0.50′′ 24 23 36.91 212.4 2549.4 8.31

OD = outside diameter and ID = inside diameter.

Lateral Load due to River Velocity

Depth = 12 ft

P = wCd × 𝛾(v)2

2(g)

Units in psf

where P = pressure in plf of pile

w = width of pile including potential debris buildup

𝛾 = unit weight of water (pcf)

𝜈 = velocity of flowing water (ft/s)

g = acceleration of gravity at Earth’s atmosphere = 32.2 ft∕s2Cd = drag coefficient for shape of pile (assume 1.3 for round)

P = 2 ft × 1.3 × 62.4 pcf9 ft∕s2

2(32.3 ft∕s2)= 204 plf


P = (2 ft) (1.3) 62.4 pcf[(9 ft∕s)2]

2(32.2 ft∕s2)= 204 plf of pile

R = 204 plf × 12 ft = 2448 lb (Figure 12.36)

FIGURE 12.36 Resultant force from water velocity.

Another method for stream velocity from the AASHTO (American Association of

State Highway and Transportation Officials) version is

P = Kv2

where K = 0.7 for round pile

K = 1.4 for HP pile

𝜈 = velocity of flowing water (ft/s)

With this formula, a velocity of 12 ft/s against an H pile would generate a 202 psf

force.

P = (1.4)(12 ft∕s2) = 202 psf

Pile can be sized for several conditions. Typically, the size would be determined

from bending or buckling since these are more critical than normal compression

stresses.

Bending from lateral 2% load and river velocity is

Mmax = (4568 lb × 12 ft) + (2448 lb × 6 ft) = 69,504 ft-lb

The required section modulus = 69,504 ft-lb × 12′′∕ft∕21,600 psi = 38.6 in3

Fb = 0.60Fy 0.60(36 ksi) = 21.6 ksi

The smallest pipe on Table 12.8 would be adequate for this bending condition

(60 in3, 18′′ × 0.25′′). This extra section modulus may be needed when combined

stress is checked.


Buckling from axial load would then be checked.

P = 229 k Try 18′′ diameter × 0.25 thick pipe (see Table 12.8)

F = P∕A F = 0.90Fy, Pall = 0.90(36 ksi) × 13.94 in2 = 451,656 lb

Due to the pile being braced, a k value of 1.0 was assumed. See Chapter 2 for end

connections.

A combined stress analysis should be looked at to make sure the pile does not fail

due to the bending and buckling combined.

Combined Stress Check

fcs = bending stress + axial

buckling stress≤ 1.0

228,405 lb

451,656 lb+ 38.6 in3

60 in3≤ 1.0

0.5057 + 0.643 ≤ 1.149, since the combined stress ratio is greater than 1.0, this com-

bined condition is exceeded by 14.9%. The next largest pile size should be considered.

The Golden State Bridge case study is a good example of a typical trestle used

over a river.

CASE STUDY: GOLDEN STATE BRIDGE—ORD FERRY BRIDGE

RETROFIT PROJECT

Project Overview

In 2011, the Golden State Bridge (GSB) of Benicia, California, was awarded

a bridge retrofit project for the County of Butte. The bridge was the Ord Ferry

Bridge over a very sensitive portion of the Sacramento River in Ord Bend, Cal-

ifornia. This project involved increasing the foundation capacity by widening

and deepening the existing foundations in the river and in the floodplain. Addi-

tional piles were added in the widened footing area. In order to accomplish this

foundation work, cofferdams were constructed around the existing footing, just

outside the new work limits and outside the existing seal course. There was

other work required to the superstructure: however, the focus of this case study

is the work access from a trestle over the river. The construction season was

short so GBS had to retrofit the east side in the first year, pull out of the river,

and then retrofit the west side the second season (Figure 12.CS4).


FIGURE 12.CS4 First season trestle—south side.

The work trestle was shown in the contract plans, but GBS was respon-

sible for their own design. They selected a trestle design they had used on

previous projects and were able to draw from this experience and do prefab-

rication of materials in one of their nearby yards. This proved to be the right

choice. The trestle design was a fairly typical design utilizing timber decking,

wide-flange(WF) stringers teamed together with steel cross bracing, and steel

WF cap beams on “star pile” (combination WF beam and T section). The river

in this area was not very deep (some areas only 4–6 ft deep). However, regard-

less of the depth, the contractor was still working over a flowing, sensitive river,

and floating equipment for access was not an option.

Crane Capacity: 110-Ton Crane, Heaviest Pick 45 k During TrestleConstruction

The trestle was designed to support a 110-ton capacity crawler crane while

hoisting a maximum load of almost 23 tons. The heaviest load the crane would

lift was not during the cofferdam construction or concrete placement. The

heaviest lift would actually be during the construction of the trestle itself.

The weight of the required boom length, rigging, hammer, leads, and pile

proved to be the most weight the crane would experience during the life of the

trestle.

(continued)



RETROFIT PROJECT (Continued)

Star Pile: Noise and Drivability in Hard Driving and Cobbles

The star pile consisted of two T sections welded to the middle of the WF beam

(Figure 12.CS5). This created a symmetrical section that had two benefits. The

first benefit was the reduction in noise and vibration. Since the Sacramento

River is habitat to a variety of fish wildlife, keeping the noise and underwater

vibrations to a minimum was critical according to the special provisions and

project permitting through the Department of Fish and Game. The second ben-

efit the star pile offered was the stout section made it easier to penetrate hard

soils and maneuver around large cobbles present in the river subsurface.

FIGURE 12.CS5 Star pile.

Teamed Beams

The stringers for the trestle arrived on the project with the cross bracing (flange

buckling resistance) and stiffeners (web yielding resistance) already installed.

This welding and other fabrication was performed in their yard, where the con-

dition for this type of work was ideal (Figure 12.CS6).

Prefabricated Cap Beams with Stiffeners and Pile Cap Cans and Pins

Also, prefabricated in the yard were the cap beams with stiffeners and pile

cap cans (Figure 12.CS7). The stiffeners were located over the pile, and the


cans were used to go over the star pile so that quick placement of the caps

could be achieved. A shear pin was placed through the can and the star pile for

uplift control. By using ideas like the prefabricated cans, welding and cutting

is eliminated, thus reducing hundreds, if not thousands, of man-hours.

FIGURE 12.CS6 Teamed stringers.

FIGURE 12.CS7 Prefabricated pile caps.(continued)



RETROFIT PROJECT (Continued)

The trestle used on this project was very common except for the use of

the star pile. To determine if the star pile concept is economical for a specific

project, one would have to do an analysis on production labor, equipment, and

materials. This same trestle concept has been used successfully throughout the

nation on various projects with pipe, H, and WF piles. The project characteris-

tics will dictate the pile types necessary to provide safe and economical access.

12.1.8 Star Pile Design and Properties

After seeing this project and the use of the star pile, this author was curious aboutthe new properties of this star pile shape. In other words, for bearing and bucklingresistance, what was the cross-sectional area and new radius of gyration? Figure 12.37shows the cross section of the star pile. What axis was now the weak axis? What wasthe shape’s moment of inertia in both the x and y axes?

Beam W18 × 97, A = 28.5 in2, Ix = 1750, Iy = 201

T’s WT6 × 26.5, A = 7.78 in2, tw = 0.345′′, d = 6.03′′, fb = 10′′, ft = 0.575,Ix = 17.7, Iy = 47.9

Calculations determined the properties of a sample star pile shown in Table 12.8.

√rx = square root of

1845.8 in4

28.5 + 2(7.78)= 6.47 in

√ry = square root of

668.6 in4

28.5 + 2(7.78)= 3.90 in

If these values are compared to a W18 × 97 by itself, the differences areinteresting.

W18 × 97 before rx = 7.82 in, ry = 2.65 in

W18 × 97 after rx = 6.47, ry = 3.90 in

The radius of gyration improved in the weak direction and was lowered in thestrong direction.

FIGURE 12.37 Cross section of star pile.

12.2 OTHER PROJECTS UTILIZING METHODS OF ACCESS 341

TABLE 12.8 Properties of Sample Star Pile

Section

Area

(in2)

d(in)

Ad2(in3)

Moment of

Inertia (in4)

ΣAd2(in4)

ΣI + Ad2(in4)

x Axis (Strong)

W18 × 97 28.5 0 0 1750 0 1750

WT6 × 26.5 7.78 0 0 47.9 0 47.9

WT6 × 26.5 7.78 0 0 47.9 0 47.9

Totals 0 1845.8 in4

y Axis (Strong)

W18 × 97 28.5 0 0 201 201

WT6 × 26.5 7.78 5.28 216.1 17.7 216.1 233.8

WT6 × 26.5 7.78 5.28 216.1 17.7 216.1 233.8

Totals 668.6 in4

12.2 OTHER PROJECTS UTILIZING METHODS OF ACCESS

The following projects are other examples of work access situations that were solved

using various methods.

The Galena Creek Project in Carson City, Nevada, shown in Figure 12.38, used

an earthen fill to raise the valley floor to an elevation that allowed the use of more

standard falsework. The cost ofmoving thematerial compacted in place and removing

the material was compared to the cost to use an alternate method of support. Alternate

methods were to cast the arch segmentally or install falsework the additional height

to the bottom of the valley floor, which was more than 200 ft.

FIGURE 12.38 Galena Creek bridge.


FIGURE 12.39 Trans Bay Terminal project.

The bridge was a combination of a flat deck supported by an arch structure. The

arch was constructed on pipe falsework and the flat deck was supported by falsework

to the completed arch.

The TransBay Terminal project in San Francisco, California, is a public trans-

portation project that is creating a terminal in downtown to merge all the major

public transportations into one terminal. The excavation takes up almost three city

blocks. Figure 12.39 shows strutting and a road crossing. The excavation included

over 3 million cubic yards of earth removal.

This aerial photo (Figure 12.40) from Google Earth shows the excavation support

system consisting of deep soil mixing and pipe strutting. Strutting was utilized despite

FIGURE 12.40 Aerial view of TransBay Terminal.

12.3 CONCLUSION 343

the excavation being over 150 ft wide. Included in the system were trestles for both

public traffic (1st St. and Fremont St.) and crane access (longitudinally) throughout

the excavation.

The very tight deflection requirements set by the engineer of record were up to

l/1000, which is 3 times most normal limits of temporary structure design. These

requirements were enforced due to the close proximity of traffic and buildings to

project site.

12.3 CONCLUSION

Access can be the difference between a successful project and an unsuccessful one.

If the contractor was worried about spending money on good access, he may not real-

ize the dividends that may pay off in the end. Sometimes it is hard to put a number on

good access and how the supporting operations may benefit. However, terrible access

will definitely not give the contractor the opportunity to maximize the production

units of all the operations relying on the access.

As with most temporary structures, without a detailed analysis (estimate), one

cannot make the most economical decisions that benefit all operations including the

subcontractors.

CHAPTER 13

SUPPORT OF EXISTING STRUCTURES

Temporary supports are used in construction when an existing structure requires tem-

porary support during demolition, construction, or retrofitting. The owner of the struc-

ture could be concerned about the structure settling, shifting, or completely failing.

Sometimes the structures continue to be occupied, such as a building or bridge; and

sometimes the facility is completely put out of service during construction. Regard-

less of the situation, the condition of the structure and safety of the public is most

important.

During the construction of temporary supports, and while the support is in service,

settlement monitoring is commonly performed to detect any unwanted settlement

or lateral movement in excess of the specifications and allowable stresses to the

structure.

During the planning stages for a temporary support design, the contractor must

consider, at a minimum, the following:

1. How stiff is the existing structure compared to the stiffness of the temporary

support? Can the existing structure allow any deflection or must the temporary

support be designed for no deflection?

2. How much does the existing structure weigh?

3. Which portion of the structure footprint will require support?

4. Where can the loads be transferred?

5. Is there sufficient support capability adjacent to the structure?

6. Are temporary piles required?

7. What materials are available?

344

13.1 BASIC BUILDING MATERIALS 345

8. Do the specifications dictate the testing methods for welding and materialusage?

9. Is the structure itself, in part, going to be part of the support? And if so, willit be able to stay within its allowable stresses?

10. Does the temporary support system have to be preloaded with a jacking deviceto eliminate or minimize deflection and stress? If so, how much preload isrequired?

Temporary supports could be necessary on a variety of project types, including thefollowing:

1. State and federal highway projects during retrofit work for seismic upgrades.

2. Building construction in congested areas to prevent undermining and settle-ment.

3. Support of utilities during crossing utility installation or junction structureconstruction.

These are just a few situations that could require temporary supports.

13.1 BASIC BUILDING MATERIALS

Structures that may need to be supported usually contain basic building materials thata contractor utilizes every day, including the following:

Concrete (150 pcf)

Reinforced concrete (160 pcf)

Steel (490 pcf or 0.2836 lb per cubic inch)

Wood with 19% moisture content (33–40 pcf)

Water (62.4 pcf)

Other materials that a contractor may need to acquire unit weights for are:

Drywall

Roofing

Siding

Fiberglass

Material weights are a function of the volume of the material and its unit weight.If a material is a mixture or composite, the conservative approach would be to usethe unit weight of the heaviest material and multiply that times the volume. Thisalmost always results in a heavier load than actual, but this is the safest approach. Ifthe volumes of the two different materials can be determined, then a more accurateweight can be determined. Another way to be more accurate and not too conservativeis to weigh the material and a “new” unit weight can be used.

Table 13.1 lists example calculations of three different materials.

346 SUPPORT OF EXISTING STRUCTURES

TABLE 13.1 Sample Calculations for Different Materials

Item L W H Volume Unit Weight Weight (lb)

Unit of 16′ long 2 × 4 16′ 4′ 3′ 192 CF 35.0 6750

Steel plate 12 6 11∕2′′ 15, 552 in3 0.2836 4,401

Concrete deadman 4 3 3 36 CF 150.0 5,400

13.1.1 Example 13.1 Pipe Unit Weight

Not all material volumes are perfect rectangles. Many times, the engineer will have touse averages, interpolate curves, round up, or subtract out voids. Pipe calculations area good example of having to subtract voids. Generally, the unit weights of pipes canbe obtained from the manufacturer’s data sheets. However, when this information isnot available, an engineer must draw from the basics in geometry and physics. Howwould the weight of a steel pipe be figured? Let’s look at how this might be calculated.

Pipe: 48′′ × 0.50′′ steel pipe. Large steel pipe is designated by its outside diameter,for example, OD = 48′′.

Contents: 1∕2 full of water.

Step 1: Calculate the outside area in inches squared because the unit weight ofsteel is in cubic inches.

A = 𝜋D2∕4

𝜋 × (48′′)2∕4 = 1810 in2

Step 2: Calculate the inside area of the pipe the same way.

𝜋 × (47′′)2∕4 = 1735 in2

Step 3: Determine the steel area by subtracting the OD from the ID as inFigure 13.1:

1810 − 1735 = 75 in2

Step 4: Calculate the weight of the pipe per linear foot:

75 in2 × 12′′(length of one foot) × 0.2836 lb∕in3 = 255 plf of pipe

FIGURE 13.1 Steel pipe example.


Step 5: Use the inside diameter in step 2 and determine the water volume if the

pipe is half full.

1735 in2

144 in2∕ft2= 12.05 ft2 × 1 ft × 1

2pipe = 6.02 ft3

Step 6: Determine the water weight per linear foot of pipe.

6.02 ft3 × 62.4 pcf = 376 plf

Step 7: What is the total weight per linear foot of this pipe, half full of water?

255 plf + 376 plf = 631 plf

These steps can be followed for almost any volume that involves an outer and an

inner portion. The unit weights of the material must be known and the dimensions

should be accessible and rounded to ensure the safest scenario.

The rest of this chapter provides examples and case studies of temporary support

of existing structures. The examples discussed may not be everyday occurrences but

will set the foundation for an engineer/designer for approaching almost any temporary

support condition, analyzing it, and making sound engineering decisions that leave

the owner confident with the plan. The examples and case studies involve a water

treatment plant building support adjacent to an excavation and a pipe support for the

construction of a junction structure.

13.1.2 Example 13.2 Existing Water Treatment Plant

A water treatment project was undertaken to increase the plant’s capacity. Part of the

scope of the work was to add a tank next to an existing building/structure. The close

proximity and depth of the new structure made it necessary to excavate adjacent to

the existing structure and risk undermining and subsequently damaging the structure.

Figure 13.2 shows the temporary structure in place. The materials being used are

steel wide-flange beams (W shapes), 25-k per leg shoring towers (legs are doubled for

50-k capacity), double steel C-channels, high-strength coil rod tension anchors, and

associated plates and nuts. The designers, for these cases, must make sound judgment

decisions when determining how much the soil underneath the structure will help

support or if the soil support should be completely neglected during the design of the

temporary support. Many structures within water and wastewater treatment plants

have to stay in service during the installation and life of the temporary support. This

means water may have to remain within the structure and its weight added to the

structure weight.

The steps necessary for temporary structure supports can follow this simple list:

1. Determine (calculate) the weight of the structure being supported, including

any anticipated live loads as well. In this case, part of the structure could be

filled with water. See Figures 13.3 and 13.4 for schematics of the structure

requiring support.


FIGURE 13.2 Temporary support of an existing structure.

FIGURE 13.3 Existing structure schematic drawing.

2. Determine where the structure can be supported. Often these locations can

experience high concentrated loads; therefore, the designer must determine

what has to happen to the structure to be able to handle the load concentra-

tion. Figure 13.3 contains dimensions to assist in weight calculations and loca-

tions for support members. This example does not measure the stiffness of the


FIGURE 13.4 Water occupying the center cell.

reinforced concrete structure. It is assumed that the structure will be supportedat enough points that stiffness is not a concern.

3. Determine what beams can be used and what are the longest spans. The projecthas W24 × 131’s, HP 14 × 89’s, and some random, smaller beams available.

Determining Structure Weight Step 1: Determine the weight of the existingstructure and associated live loads.

Front Wall: (25 × 11 × 12′′) × 160 pcf = 44, 000 lb

Wall 1 Ext (2 ea): 2(7 × 11 × 12′′) × 160 pcf = 24, 640 lb

Wall 2 Int (2 ea): 2(7 × 11 × 2′′) × 160 pcf = 24, 640 lb

Slab on grade: (25 × 7 × 12′′) × 160 pcf = 44, 000 lb

Water in center cell: 6 × 7 × 8 × 62.4 pcf = 20, 966 lb

Total structure weight = 158, 246 lb (160 k)

Weight per linear foot of building (average) 158.3 kips∕25 ft = 6.4 klf

The layout was conducive to using double stringers. These stringers would besupported by shoring towers on the ground on one side and the existing portion ofthe structure founded on a lower, stable level on the other side. Figure 13.5 shows theplan view of this temporary support layout.

FIGURE 13.5 Plan view of temporary support layout.


FIGURE 13.6 Two loading condition options.

The load can be figured as a linear load (6.4 klf) or as three concentrated loadsrepresenting each cell (see Figure 13.6). Regardless of the method, the maximummoment figured should be the worst-case scenario.

Method 1: 6.4 klf (see previous calculation).Method 2: Four concentrated loads, one from each wall of the three cells. The

center cell is heavier due to the water inside the channel.Method 1 from Beam Software:

Mmax = 1140 ft-k

Vmax = 80 k

Method 2 from Beam Software:

Mmax = 1087 ft-k

Vmax = 82 k (loads off center slightly)

Stringer Design Using the most conservative method would mean using themoment and shear from method 1 shown in Figure 13.7. With these values we cancheck if twoW24 × 131 work side by side. For clarity, the weight of the actual beamswill not be added to the live loads. In practice, these weights would not be ignored.

Allowable stresses for A36 steel:

Fb = 0.60Fy = 21.6 ksi

Fv = 0.40Fy = 14.4 ksi

FIGURE 13.7 Shear and moment diagram method 1.


Sx for a single W24 × 131 = 329 in3, two beams Sx = 329 in3 × 2 = 658 in3

Using method 1:

fb = (1140 ft-k × 12′′∕ft)∕658 in3 = 20.8 ksi < 21.6 ksi (OK)

fv = 82 k∕2(24.48′′ × 0.605′′) = 2.77 ksi < 14.4 ksi (OK)

Check the stability of the beam with d∕Af = 1.98.20, 000∕(36 × 1.98)∕12′′∕ft = 24.0 < 41 ft, brace one beam to the other at least

in the middle. Two braces would allow the contractor to set the beams as a team andoffer more stability.

Cap Beam Design The reaction at the tower from the previous beam designis 82 k. This is the force from the double stringer to the HP14 × 89 cap beam.Figure 13.8 shows the tower arrangement.

The design of theHP14 × 89 cap beam has a double point load from the twoW24 ×131’s on the HP14 × 89 spanning from one side of the tower to the other side. Thetwo point loads can be combined as one 82-k load in order to simplify the problemand still end up with the same moment. The tower is made up of double frames oneach side. They are generally 25 k per leg frames, but, when doubled up, become 50 kper leg towers. The span of the HP14 × 89 is approximately 6 ft. Therefore, using the82-k load and a 6-ft span, calculate the maximum bending moment in the cap beamas shown in Figure 13.9. As a rule, the shear should also be checked.

M = PL∕4

M = 82 k (6 ft)∕4 = 123 ft-k

Fb = 123 ft-k × 12′′∕ft∕131 in3 = 11.27 ksi < 21.6 ksi (OK)

Fv = 41 k∕(13.83′′ × 0.615′′) = 4.82 ksi < 14.4 ksi (OK)

Subcap Beam Design The subcaps run the short distance of the tower (4 ft),and there are two on each side in order to distribute the cap beam load equally.

FIGURE 13.8 Tower arrangement supporting stringers.


FIGURE 13.9 Cap beam layout, end view section.

FIGURE 13.10 Subcap beam layout.

The company has available some short W8 × 31 × 8′ beams. Determine if this size

beam will be sufficient. Apply the force from the cap beam to the center of the subcap

beam similar to Figure 13.10. There are double subcaps on each side so be sure to

divide the total 82-k force by 4.

Properties of a W8 × 31:

Sx = 27.5 in3

d = 8.0′′

tw = 0.285′′

M = PL4

M =(82 k∕4 ea)(4 ft)

4= 20.5 ft-k

fb =(20.5 ft-k)(12′′∕ft)

27.5 in3= 8.95 ksi < 21.6 ksi


This is more than half the allowable stress. Would one subcap beam work?

fb =41 ft-k × 12′′∕ft

27.5 in3= 17.9 ksi < 21.6 ksi (OK)

A single subcap beam would work. However, the reason a contractor may have

used the double tower (25 × 2 = 50 k) is to be sure the tower capacity is adequate.

The next few steps will address the towers.

The tower leg loads come from the total 82-k force to the whole tower. If this 82-k

force was distributed equally to the eight tower legs, the force on each leg would be

slightly over 10 k. In this case, this falsework system is slightly underused.

82 k

8 legs= 10.25 k∕leg < 25 k ( OK)

The foundation for this system could be a series of timber pads distributing the

tower leg forces to the soil. The pads must be a safe distance from the top of slope

into the excavation. Figure 13.11 shows the location of the tower supports and the

importance of this distance. A pair of two legs are supporting 20.5 k in this example.

Figure 13.11 shows four 4 × 6’s supporting a 4 × 6 corbel. If the 4 × 6 corbel projects

a 1 :1 load path to the supporting soil, what would the soil load be?

F = 20.5 k∕{(4 × 5.5′′) × [5.5′′ + (2 × 3.5)]}∕144 in2 = 10.73 ksf or 10, 730 psf

Besides some rare occasions, soil pressure can support anywhere from 2000 psf to

6000 psf. Therefore, this soil either needs to be supported on imported aggregate fill

compacted to 95% or the base dimensions on the corbels and pads need to be widened.

FIGURE 13.11 Elevation view of structure.


FIGURE 13.12 Pad load distribution.

FIGURE 13.13 Corbel and pad (sill) arrangement.

The projection of the load through the timber pads is illustrated in Figure 13.12.Figure 13.13 shows the actual design condition of the corbels and pads (sills).

The two beams are attached to the structure with coil rods, plates, and nuts (seeFigure 13.14). The coil rod capacity should be adequate to resist the weight of theportion of structure associated with the rod location. In this case, four rods are pro-posed. If we go back to the method 2 case for the stringer design, the two center rodswere estimated to hold 30 k at the two outside rods and 50 k at the two inside rods. Ifthe system was designed to support the 50-k load, then the other rods would be morethan adequate. Of course, the two scenarios could have different rod diameters, butthen the project staff would have to inventory these rods, and the chance of placingthe wrong rod in the wrong location would increase.

Appendix 8 shows coil rod capacities. The Williams coil rod that has minimumyield strength of 58.1 k is the 1

1

4′′ diameter coil rod.

13.1.3 Example 13.3 Temporary Pipe Supports

The temporary support example that will be illustrated next is the support of an exist-ing pipe that will be undermined in order to build a complete new structure around it


FIGURE 13.14 Coil rod attachment.

FIGURE 13.15 Pipe support.

(see Figure 13.15). This condition is fairly common when a water facility desires to

direct water to multiple locations. By adding a junction box, the operators can direct

flow to one, two, or three directions or stop the flow completely. The pipe in this

case may have to remain in service so the plant can continue water delivery. It is not

until the junction structure is complete and the pipe inside demolished that the new

structure will operate with its new configuration.

The pipe is assumed to be 50% filled with water during the life of the tempo-

rary support. The support system should be incorporated with the shoring of the

excavation. If there is no shoring and the excavation is sloped (open-cut), then the


FIGURE 13.16 Pipe and support structure.

increased plan dimension of the excavation should be considered. An increased plandimension would significantly lengthen the required beams.

The pipe that needs to be supported is an 84-in (inside diameter) reinforced con-crete pipe (RCP). The water weight inside the pipe will be assumed at 62.4 pcf. Pipeweights can usually be looked up in a manufacturer’s table. However, this pipe hasbeen in the ground for 18 years and the specifications could not be found. Therefore,the unit weight of water, concrete, and steel with the dimensions of the pipe will beused to determine the weight per linear foot of pipe. The excavation is 52 ft acrossfrom one shoring wall to a steel frame support at the other side. Three stringers span-ning the 52-ft excavation will be used due to a slight curve in the pipe. The generalarrangement of the pipe and support structure are shown in Figure 13.16. The pipe isslightly skewed to the new structure design. Unlike Example 13.2, the weight of thecomponents will be carried to the next step as the problem is solved.

The load path for supporting the RCP will follow this path:

Pipe will be saddled by wire rope.

The wire rope will terminate through a double C-channel support over the threestringers using coil rod, eye nuts, and wire rope clips.

The stringers will span 52 ft from one side of the excavation to the other.

The stringers will be supported at each end with the shoring wall or a short capbeam.

This example will follow the described load path using a series of steps. StandardA36 steel will be assumed using a 0.60Fy factor of safety.

Weight of Existing Conditions Step 1: Determine the weight of permanent pipecontaining 50% water.

Reinforced Concrete Pipe RCP sizes are designated by the inside diameter (ID).Therefore, an 84′′ RCP measures 84′′ inside, and the outside dimension (OD) is thesum of the ID plus twice the pipe’s thickness. The 84′′ RCP is approximately 8′′

thick, which means the OD is 84′′ + (2 × 8′′) = 100′′. The process of calculatingpipe weight at the beginning of this chapter holds true for concrete as well. The unitweight of the material becomes the variable:

OD =𝜋(100′′∕12′′∕ft)2

4= 54.5 SF

ID =𝜋(84′′∕12′′∕ft)2

4= 38.5 SF

Concrete area = 54.5–38.5 = 16 SF × 1 ft = 16 CF

Pipe weight = 16 SF × 160 pcf = 2560 lb pipe weight per linear foot


Water The pipe can only be half full at its maximum operating capacity due tothe hydraulics of the plant. This must be confirmed so that, during the life of thistemporary support, the pipe will never contain more than 50% water.

Using the ID from the pipe above, the weight of 50% water can be obtained.

ID = 38.5 SF

Water volume = 38.5 SF × 1 ft = 38.5 CF

Water weight = 38.5 CF × 62.4 pcf = 2402 lb × 50% = 1201 lb water weightper LF

Total weight of pipe and water per liner foot = 3761 lb

Step 2: Determine the size required for the wire rope.At this point, it should be decided how far apart the wire rope saddles and double

channels will be spaced. If there was a particular cable and channel size available,the engineer could use reverse calculations to obtain the proper spacing to match thecable and channel available. Another approach would be to choose spacing for thewire rope and channels, then size the cable and choose the channel. Moving forward,the spacing will be assumed to be 6 ft on center from saddle to saddle. Using thisspacing, determine the weight per saddle and size the wire rope. The wire rope asa saddle is considered a two-parted line similar to a two-parted crane line that goesfrom the boom tip, through a “headache ball,” and back to the boom with a wedgeand socket (becket). In other words, the load that is distributed from the pipe to thewire rope is shared by the two legs of the wire rope.

The load associated with each saddle support is the weight of the pipe and watertimes 6 ft of pipe (6 ft on center).P = 3761 lb × 6 ft = 22, 566 lb∕2 parts of line = 11, 283 lb on a single wire rope

as shown in Figure 13.17At this time since this figure will dictate the rest of the design, it is appropriate to

round this value to the nearest 100 lb and into kip units, say 11.3 k.The factor of safety on wire rope can be between 2.0 and 6.0 (Table 13.2), depend-

ing on the application. For this application, the engineer should consider the risk in

FIGURE 13.17 Wire rope saddle support.


TABLE 13.2 Wire Rope Safe Working Loads (SWL)

Minimum Breaking Force and Allowable Loads for 6 × 19 IWRC-XIPS Wire Rope with

Wire Rope Reduction of 80% < 1′′ and 90% 1′′ and Greater for Using Wire Rope Clips

Wire Rope

Diameter

(in)

Breaking

Strength

(tons)

SWL w/

2.0 FOS &

80–90%

(tons)

SWL w/

3.0 FOS &

80–90%

(tons)

SWL w/

4.0 FOS &

80–90%

(tons)

SWL w/

5.0 FOS &

80–90%

(tons)

SWL w/

6.0 FOS &

80–90%

(tons)

3

8—80% 7.6 3.0 2.0 1.5 1.2 1.0

1

2—80% 13.3 5.3 3.5 2.7 2.1 1.8

5

8—80% 20.6 8.2 5.5 4.1 3.3 2.7

3

4—80% 29.4 11.8 7.8 5.9 4.7 3.9

7

8—80% 39.8 15.9 10.6 8.0 6.4 5.3

1—90% 51.7 23.3 15.5 11.6 9.3 7.8

this pipe moving, shifting, or failing. The consequences, if any of these misfortunesoccur, could be very costly to the contractor and probably shut down the existingplant, thus costing the owner lost revenue. With this said, the contractor will use a4:1 factor of safety from the ultimate strength of the wire rope capacity.

P × 4.0 = safe working load (SWL)

11.3 k = 5.7 tons

Referencing thewire rope chart shown in Table 13.2with a 4 :1 FOS and usingwirerope clips, the minimum wire rope size that can be used is 3

4′′ 6 × 19 IWRC-XIPS.

To transition from wire rope to the tops of the double channel, the following hard-ware is required:

Wire rope clips

Eye bolt for the wire rope to loop through

Thimble to soften the radius of the wire rope bent

Coil rod threaded into the wire rope and going through the double channel

Plate washer (5 × 5)3

4′′ nut and washer

The hardware charts in Appendix 8 will indicate the SWL for each component. Allthe hardware associated with 3

4′′ diameter coil rod has an SWL of 19,000 lb, which

includes a manufacturer recommended 2:1 FOS. Therefore, the 3

4′′ hardware with a

5 × 5 × 1

2plate is adequate.

Double Channel Selection Step 3: Extend the wire rope saddles through thedouble channels and determine the most economical channels necessary to supportthe intended load.


FIGURE 13.18 Double channels loading stringers.

The 11.3-k load will be applied to the channel in two locations and the channels

will be supported by three stringers. A free-body diagram should be drawn similar

to Figure 13.18. The three reactions at the bottom represent the three stringers, and

the two forces coming from above represent the wire rope forces connected to the

double channels. The wire rope connections are exactly 100 in apart because that is

the outside diameter of the pipe and the cable wraps around the pipe. The stringers

are placed 6 ft apart from each other.

Since the pipe is skewed, the engineer is looking for the worst-case scenario that

causes the largest bending moment to the double channel. As it turns out, the most

symmetrical arrangement also causes the largest bendingmoment. The beam program

is used in Figure 13.19 because of the indeterminate loading condition. See the results

below for the double channel.

Mmax = 11.5 ft-k

Vmax = 6.29 k

Stringer reactions:

Left: 6.3 k

Center: 10.0 k

Right: 6.3 k

FIGURE 13.19 Shear and moment diagram of double channel.


With this information, the double channel size can be determined. These reactionswill become the point loads on the stringers for the sizing of the stringers as well.The minimum size channel that can be used to resist a maximum bending momentrequires an Sx = 11.5 ft-k × 12′′∕ft ∕21.6 ksi = 6.4 in3∕2 channels = 3.2 in3 for asingle channel. A C5 × 9 has the closest section modulus that works; however, aC6 × 8.2 is lighter (9 lb vs. 8.2 lb). Either of these will work, but the 6′′ channelwould be more economical.

The shear can be checked on a channel in the same way shear is checked on awide-flange beam. The depth of the beam and the thickness of the web make upthe area resisting shear, and this is divided into the maximum shear value as such.Appendix 1 is used for d and tw.

fv = 6.29 k∕2 channels × (6′′ × 0.20′′) = 2.62 ksi < 14.4 ksi (OK)

Stringer Selection Step 4: Determine the size of all three stringers.Draw a free-body diagram of the double channels loading on each of the three

52-ft-long stringers. There are two beam cases, the outside stringers, and the middlestringer. The shear and moment diagram in Figure 13.19 indicates the loads on thestringer without the weight of the channels from the three reaction values shown:6.3, 10, and 6.3 k respectively. The downward forces represent the double channelsloading the stringers, and the reactions acting upward represent the cap beam supportson each side of the excavation. The force from the channel now includes the weightfor the channels and the wire rope hardware. With the addition of 300 lb pounds perside, the 10-k load becomes 10.3 k and the 6.3-k load on the outside beams is now6.6 k. Figure 13.20 shows nine double channel sets loading a stringer.

The middle beam was put into the beam program and generated the followingresults (shown in Figure 13.21):

Mmax = 567 ft-k

Vmax = 46.2 k

Reactions = 46.2 k

From this information, we can determine the beam size:

Sx = 568 ft-k × 12′′∕ft ∕21.6 ksi = 312 in3

The W24 × 131 beam is adequate for bending, as is the W18 × 158.The two outside beams can be much smaller. However, it is advisable to maintain

the same beam height so that the cap beam and the double channels are providedtwo level surfaces on which to rest. Therefore, W24 wide flanges can be selectedfor the outsides as well (much lighter, of course). If this is not desired because theW24 becomes too light, then the center beam can be reduced to a W18 × 158 and theoutside beam would be an 18′′ beam as well. Lateral bracing should also be analyzed.

FIGURE 13.20 Double channels loading stringer.


FIGURE 13.21 Shear and moment diagram of middle stringer.

FIGURE 13.22 Shear and moment diagram of outside stringers.

This beam arrangement is favorable to bracing because the beams are fairly close

together (6 ft apart) and parallel to each other.

Let’s determine the size of the outside beams using the same free-body diagram,

but including the lighter channel loads of 6.6 k. A new shear and moment diagram is

necessary, as shown in Figure 13.22.

Mmax = 376.2 ft-k

Vmax = 29.7 k

Reaction = 29.7 k

Sx = 376.2 ft-k × 12′′∕ft ∕21.6 ksi = 209 in3

W24 × 94 and W18 × 119 are adequate for bending.


TABLE 13.3 Cost Comparison between W18s and W24s

Nominal

Depth

Inside

Stringer (1)

Outside

Stringers (2)

Weight

(52′ long) lb

Total $

(at $0.65)

W24 option W24 × 131 W24 × 94 16,588 $10,782

W18 option W18 × 158 W18 × 119 20,592 $13,385

TABLE 13.4 Lateral Bracing Requirements

Beam d∕Af Lc (ft) Braces Cost per Brace Total Cost

W24 × 131 1.98 21.3 2 $1000 N/A

W24 × 94 3.06 13.8 3 $1000 $3000

W18 × 158 1.21 34.8 1 $1000 N/A

W18 × 119 1.59 26.5 1 $1000 $1000

The options for the three stringers (looking at only 18′′ and 24′′ WF beams) are

both combinations:

1. W24 option: W24 × 131 on the inside and W24 × 94’s (two each) on the out-

side, or

2. W18 option: W18 × 158 on the inside and W18 × 119’s (two each) on the out-

side.

Table 13.3 shows the cost analysis and Table 13.4 the bracing analysis.

Pile Caps The pile cap design is greatly affected by the number of piles supporting

the stringers. In this example, the one side is supported on a continuous sheet pile wall,

and the other side is supported on a cap beam supported by individual pile. The latter

will be analyzed for the pile cap and pile loading and sizing.

Since the piles have to be outside the limits of the pipe being supported, the beam

arrangement would be represented by the sketch shown in Figure 13.23. The loads

from the three stringers can be found from the shear and moment diagrams (shown

earlier in Figure 13.22). The stringers load the cap beam with a 29.7-k load from the

two outsides and a 46.2-k load from the inside stringer. A beam weight should be

added to both sides for the inside and outside stringers. An assumed weight of 6 k for

the outside beams and 8 k for the inside beam will be adequate. Half of each of these

additional weights will be added to the loads at the ends of the stringers.

29.7 +(1

2of 6 k

)= 32.7 k

46.2 +(1

2of 8 k

)= 50.2 k


FIGURE 13.23 Cap beam loaded by stringers.

The piles are 12-ft apart on the opposite side of the shoring wall. Figure 13.23 lays

out the loads on the cap beam supporting the stringers.

Maximum bending moment from the beam program (not shown) is 150.6 ft-k.

The beam used for this would require an 82.3-in3 section modulus. Since HP

shapes are good to use as cap beams, the lightest HP shape available is an HP 13 × 73

or an HP 14 × 73 (both weighing the same). There is a high concentrated load from

the pile to the cap beam; therefore, the cap beam will need to be checked for web

yielding. The pile size will not be determined so in order to check web yielding,

it will be assumed that the pile width will be 12 in along the beam. The shear and

moment diagram for these piles indicate a right and left reaction value of 57.8 k. This

reaction is what the pile will have to support without a safety factor. This will also be

the concentrated load used for web yielding. Aweb yielding sketch has been provided

in Figure 13.24.

The values needed in order to check web yielding are the pile width, which was

just established as 12 in, the k dimension of the cap beam(11

8′′), and the thickness

of the cap beam web (0.46 in).

FIGURE 13.24 Web yielding at pile.


The concentrated load of 57.8 k is distributed at a 2.5 :1 angle, and the yielding

stress occurs where it intersects the line at k distance. The HP 13 × 73 will be checked

since it has a thicker web:

fwy = P∕[tw × (w + 5k)]

fwy = 57.8 k∕{(0.565′′) × [12′′ + (5 × 1.25′′)]} = 5.61 ksi < 23.78 ksi (OK)

CASE STUDY: RETROFIT PROJECT IN SAN FRANCISCO,

CALIFORNIA

Project Overview

After the Loma Prieta earthquake shook northern California in 1989, a decade

of projects were initiated, including projects that widened and reinforced foot-

ings, wrapped columns in steel jackets, and replaced complete substructures.

Such projects were something of the norm from 1990 to almost 2000. Many

heavy civil contractors who did not want to miss out on an opportunity to con-

tract in a new kind of work began to bid and buildmany of these retrofit projects.

Since this work was fairly new, it brought not only new, exciting projects to the

workplace but also a significant amount of uncertainty. What used to be con-

sidered normal ways to build bridges and their components was being changed.

Even the California Department of Transportation had to develop standards and

best practices for contractors to follow. The welding code previously used for

temporary structures was changed to includewelding procedures normally used

for permanent structures. Figure 13.CS1 shows a schematic of a typical viaduct

structure with two levels.

FIGURE 13.CS1 Viaduct schematic—two levels.

During this period, one structure of concern (following the collapse of a

two-story bridge in Oakland) was a viaduct that spanned Silver Ave and High-

way 101 and merged onto the Highway 280 corridor in San Francisco. This

viaduct, originally constructed in the early 1960s, had sections that were con-

stant but also had sections that were unique. Figure 13.CS2 shows a portion of

Highway 280 going over Highway 101. The main purpose of this project was


to support the existing reinforced concrete box-girder bridge; remove the foot-

ings, columns, and portions of the bent caps; install new piles, columns, and

bent caps; and construct a longitudinal edge beam along each side of the lower

deck portion. The project was sequenced (by specification) so that no adjacent

bents could be worked on within a single frame. A frame on this project (hinge

to hinge) consisted of approximately three to four bent caps. The bridge in most

cases was two levels, and the lanes on the two levels were mostly closed to traf-

fic for the duration of the project. In addition to the retrofitted bent caps, several

hinges were removed completely (edge of deck to edge of deck) and rebuilt with

new, state-of-the-art spherical Teflon bearings. While the hinge was removed,

the bridge had to be supported back to the closest bent cap (the short span)

while supporting the long span.

FIGURE 13.CS2 Highway 280 split going over Highway 101 in San Francisco.

Bent Cap Replacement

There were 5 frames with a total of 23 bents (approximately 4 to 5 bents per

frame). As mentioned earlier, only one bent per frame could be retrofitted at

once. In other words, only 5 bents could be worked on at the same time. Also,

2 adjacent bents could not be worked on simultaneously. In addition to these

constraints, other specification constraints made it so there were actually only

4 bents being worked on simultaneously. As the project moved forward, part-

nering efforts allowedmore work to be accomplished at the same time but never

enough to satisfy an aggressive schedule.

(continued)



CALIFORNIA (Continued)

The image shown in Figure 13.CS3 was a typical bent cap retrofit. The

bridge was supported by temporary supports on each side of the bent cap. The

temporary supports were spaced far enough apart to allow access to the demo-

lition and structures crew while at the same time not increasing the tributary

load that had to be supported. The supports consisted of WF posts, WF beams,

T shape bracing, and was founded on temporary piling in most cases. Hydraulic

jacks and short beams were used between the bottom of the bridge and the top

of the main beam. The system was jacked to a prescribed load, and then beam

spacers were placed until the system was de-stressed and removed.

FIGURE 13.CS3 Typical bent cap replacement.

The demolition consisted of concrete removal where shown on the draw-

ings and in almost all cases required the reinforcing steel to remain. The limits

of box-girder removal were limited to 2 ft on each side of the bent cap, which

would be replaced by additional new bent cap concrete (8 ft). The original bent

caps were 4 ft wide. The reinforcing bar placement was difficult and couplers

were used extensively to join the new bars to the old, remaining bars. In new

construction, the ironworkers can determine the order of bar placement in the

most economical fashion. In retrofit construction, where existing bars remain,

the order of placement is somewhat dictated by the remaining bars. It should

also be pointed out that every operation had to work around the temporary sup-

ports on each side of the bent.


Welding was critical on this project. The owner specified very strict welding

requirements that would normally be used for new construction, and welder

certifications were time consuming and expensive. Most of the welding was

performed on-site so it took a while to get the on-site welding operations to run

as smoothly as a shop might perform. In most cases, this goal was not reached.

Hinge Replacement

During the hinge replacement, the complete hinge, including concrete and

reinforcing steel (except approximately 4 ft of reinforcing on each side), was

removed. The bridge was approximately 40 ft wide and about 20 ft of hinge

was removed longitudinally. The temporary support system selected doubled

as a demolition platform and as support for the new concrete to be placed.

The above deck support beams spanned from one bent to close to another

bent since the removal took away any possibility that the bridge itself could

offer any support. The work on the new hinge took approximately 2 months

until the temporary support could be removed. Figure 13.CS4 shows a hinge

support in place and the existing hinge concrete removed.

FIGURE 13.CS4 Hinge support.

The above deck beams were W36 beams and were capped at the locations

where the support rods went through the deck and were also diagonally braced.

(continued)



CALIFORNIA (Continued)

The support rods went through holes that were drilled into the top and bottom

deck of the box girder and terminated under the bridge with plates and nuts. The

rods were 11

4′′ and 1

1

2′′ Dywidag type with 150-ksi high-strength capacity.

The falsework deck/demo platform was also supported by these rods. The

new hinge was built in four stages of concrete placements and included the

new bearing installation. The demolition was done either with small machines

or by hand to preserve the remaining concrete and reinforcing steel. The spec-

ifications were very strict on not damaging the reinforcing steel that was to

remain; and if damage did occur, the bars had to be fill-welded with welding

rod. Figure 13.CS5 is a sketch of a typical hinge support and how the loads

were transferred to each adjacent bridge bent.

FIGURE 13.CS5 Sketch of hinge replacement.

The three examples provided in this chapter were from the author’s expe-

riences and represent more challenging and unusual temporary support sit-

uations. The case study discussed may be more representative of temporary

support situations that onemay encounter in heavy civil construction.Whatever

the situation, contractors should procure the services of a qualified, registered,

licensed engineer to assist in a safe and economical design and to distribute

some unnecessary risk.

The student of temporary structure design should be proficient in calculating

weights of all materials knowing the overall dimensions and unit weights of the

materials. With these basic skills, one should be able to assist in any temporary

structure supporting existing structures, buildings, and utilities.

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APPENDIXES

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Trim Size: 6.125in x 9.25in Souder bapp01.tex V2 - 09/12/2014 4:10pm Page 371

APPENDIX 1

STEEL BEAMS (AISC)

371


Wide-Flang

eBeams

Designation

Area

ADepth

d

Web

Flange

Lateral

Buckling

Axisx-x

Axisy-y

Thickness

t w

Width

b f

Thickness

t fk

d/Af

I xS x

r xI y

S yr y

(in2)

(in)

(in)

(in)

(in)

(in)

(in4)

(in3)

(in)

(in4)

(in3)

(in)

W44×3

35

98.3

44.02

1.02

15.95

1.77

29/16

1.559

31100

1410

17.8

1200

150

3.49

W44×2

90

85.8

43.62

0.87

15.83

1.58

23/8

1.744

27100

1240

17.8

1050

133

3.5

W44×2

62

77.2

43.31

0.79

15.75

1.42

23/16

1.937

24200

1120

17.7

927

118

3.46

W44×2

30

67.7

42.91

0.71

15.75

1.22

22.233

20800

969

17.5

796

101

3.43

W40×5

93

174

42.99

1.79

16.69

3.23

47/16

0.797

50400

2340

17

2520

302

3.81

W40×5

03

148

42.05

1.54

16.42

2.76

315/16

0.928

41700

1980

16.8

2050

250

3.72

W40×4

31

127

41.26

1.34

16.22

2.36

39/16

1.078

34800

1690

16.6

1690

208

3.65

W40×3

72

109

40.63

1.16

16.06

2.05

31/4

1.234

29600

1460

16.4

1420

177

3.6

W40×3

21

94.1

40.08

115.91

1.77

215/16

1.423

25100

1250

16.3

1190

150

3.56

W40×2

97

87.4

39.84

0.93

15.825

1.65

31/16

1.526

23200

1170

16.3

1090

138

3.54

W40×2

77

81.3

39.69

0.83

15.83

1.575

23/4

1.592

21900

1100

16.4

1040

132

3.58

W40×2

49

73.3

39.38

0.75

15.75

1.42

25/8

1.761

19500

992

16.3

926

118

3.56

W40×2

15

63.3

38.98

0.65

15.75

1.22

23/8

2.029

16700

858

16.2

796

101

3.54

W40×1

99

58.4

38.67

0.65

15.75

1.065

21/4

2.305

14900

769

16

695

88.2

3.45

W40×1

74

51.1

38.2

0.65

15.75

0.83

22.922

12200

639

15.5

541

68.8

3.26

W40×4

66

137

42.44

1.67

12.64

2.95

41/8

1.138

36300

1710

16.3

1010

160

2.72

372


W40×3

92

115

41.57

1.42

12.36

2.52

311/16

1.335

29900

1440

16.1

803

130

2.64

W40×3

31

97.6

40.79

1.22

12.17

2.13

35/16

1.574

24700

1210

15.9

646

106

2.57

W40×2

78

81.8

40.16

1.02

11.97

1.81

31.854

20500

1020

15.8

521

87.1

2.52

W40×2

64

77.6

40

0.96

11.93

1.73

215/16

1.938

19400

971

15.8

493

82.6

2.52

W40×2

35

68.9

39.69

0.83

11.89

1.575

23/4

2.119

17400

874

15.9

444

74.6

2.54

W40×2

11

62

39.37

0.75

11.81

1.415

25/8

2.356

15500

785

15.8

390

66.1

2.51

W40×1

83

53.7

38.98

0.65

11.81

1.22

23/8

2.705

13300

682

15.7

336

56.9

2.5

W40×1

67

49.1

38.59

0.65

11.81

1.025

23/16

3.188

11600

599

15.3

283

47.9

2.4

W40×1

49

43.8

38.2

0.63

11.81

0.83

23.897

9780

512

14.9

229

38.8

2.29

W36×8

48

249

42.45

2.52

18.13

4.53

511/16

0.517

67400

3170

16.4

4550

501

4.27

W36×7

98

234

41.97

2.38

17.99

4.29

57/16

0.544

62600

2980

16.4

4200

467

4.24

W36×6

50

190

40.47

1.97

17.575

3.54

411/16

0.650

48900

2420

16

3230

367

4.12

W36×5

27

154

39.21

1.61

17.22

2.91

41/16

0.782

38300

1950

15.8

2490

289

4.02

W36×4

39

128

38.26

1.36

16.965

2.44

39/16

0.924

31000

1620

15.6

1990

235

3.95

W36×3

93

115

37.8

1.22

16.83

2.2

35/16

1.021

27500

1450

15.5

1750

208

3.9

W36×3

59

105

37.4

1.12

16.73

2.01

31/8

1.112

24800

1320

15.4

1570

188

3.87

W36×3

28

96.4

37.09

1.02

16.63

1.85

31.206

22500

1210

15.3

1420

171

3.84

W36×3

00

88.3

36.74

0.945

16.655

1.68

213/16

1.313

20300

1110

15.2

1300

156

3.83

W36×2

80

82.4

36.52

0.885

16.595

1.57

211/16

1.402

18900

1030

15.1

1200

144

3.81

(continued)

373


Designation

Area

ADepth

d

Web

Flange

Lateral

Buckling

Axisx-x

Axisy-y

Thickness

t w

Width

b f

Thickness

t fk

d/Af

I xS x

r xI y

S yr y

(in2)

(in)

(in)

(in)

(in)

(in)

(in4)

(in3)

(in)

(in4)

(in3)

(in)

W36×2

60

76.5

36.26

0.84

16.55

1.44

29/16

1.521

17300

953

15

1090

132

3.78

W36×2

45

72.1

36.08

0.8

16.51

1.35

21/2

1.619

16100

895

15

1010

123

3.75

W36×2

30

67.6

35.9

0.76

16.47

1.26

23/8

1.730

15000

837

14.9

940

114

3.73

W36×2

56

75.4

37.43

0.96

12.215

1.73

25/8

1.771

16800

895

14.9

528

86.5

2.65

W36×2

32

68.1

37.12

0.87

12.12

1.57

21/2

1.951

15000

809

14.8

468

77.2

2.62

W36×2

10

61.8

36.69

0.83

12.18

1.36

25/16

2.215

13200

719

14.6

411

67.5

2.58

W36×1

94

57

36.49

0.765

12.115

1.26

23/16

2.390

12100

664

14.6

375

61.9

2.56

W36×1

82

53.6

36.33

0.725

12.075

1.18

21/8

2.550

11300

623

14.5

347

57.6

2.55

W36×1

70

50

36.17

0.68

12.03

1.1

22.733

10500

580

14.5

320

53.2

2.53

W36×1

60

47

36.01

0.65

12

1.02

115/16

2.942

9750

542

14.4

295

49.1

2.5

W36×1

50

44.2

35.85

0.625

11.975

0.94

17/8

3.185

9040

504

14.3

270

45.1

2.47

W36×1

35

39.7

35.55

0.6

11.95

0.79

111/16

3.766

7800

439

14

225

37.7

2.38

W33×3

54

104

35.55

1.16

16.1

2.09

27/8

1.056

21900

1230

14.5

1460

181

3.74

W33×3

18

93.5

35.16

1.04

15.985

1.89

211/16

1.164

19500

1110

14.4

1290

161

3.71

W33×2

91

85.6

34.84

0.96

15.905

1.73

29/16

1.266

17700

1010

14.4

1160

146

3.69

374


W33×2

63

77.4

34.53

0.87

15.805

1.57

23/8

1.392

15800

917

14.3

1030

131

3.66

W33×2

41

70.9

34.18

0.83

15.86

1.4

23/16

1.539

14200

829

14.1

932

118

3.63

W33×2

21

65

33.93

0.775

15.805

1.275

21/16

1.684

12800

757

14.1

840

106

3.59

W33×2

01

59.1

33.68

0.715

15.745

1.15

115/16

1.860

11500

684

14

749

95.2

3.56

W33×1

69

49.5

33.82

0.67

11.5

1.22

21/16

2.411

9290

549

13.7

310

53.9

2.5

W33×1

52

44.7

33.49

0.635

11.565

1.055

17/8

2.745

8160

487

13.5

273

47.2

2.47

W33×1

41

41.6

33.3

0.605

11.535

0.96

13/4

3.007

7450

448

13.4

246

42.7

2.43

W33×1

30

38.3

33.09

0.58

11.51

0.855

111/16

3.362

6710

406

13.2

218

37.9

2.39

W33×1

18

34.7

32.86

0.55

11.48

0.74

19/16

3.868

5900

359

13

187

32.6

2.32

W30×4

77

140

34.21

1.63

15.865

2.95

33/4

0.731

26100

1530

13.7

1970

249

3.75

W30×3

91

114

33.19

1.36

15.59

2.44

31/4

0.873

20700

1250

13.5

1550

198

3.68

W30×3

26

95.7

32.4

1.14

15.37

2.05

213/16

1.028

16800

1030

13.2

1240

162

3.61

W30×2

92

85.7

32.01

1.02

15.255

1.85

25/8

1.134

14900

928

13.2

1100

144

3.58

W30×2

61

76.7

31.61

0.93

15.155

1.65

27/16

1.264

13100

827

13.1

959

127

3.54

W30×2

35

69

31.3

0.83

15.055

1.5

21/4

1.386

11700

746

13

855

114

3.52

W30×2

11

62

30.94

0.775

15.105

1.315

21/8

1.558

10300

663

12.9

757

100

3.49

W30×1

91

56.1

30.68

0.71

15.04

1.185

115/16

1.721

9170

598

12.8

673

89.5

3.46

W30×1

73

50.8

30.44

0.655

14.985

1.065

17/8

1.907

8200

539

12.7

598

79.8

3.43

W30×1

48

43.5

30.67

0.65

10.48

1.18

22.480

6680

436

12.4

227

43.3

2.28

(continued)

375


Designation

Area

ADepth

d

Web

Flange

Lateral

Buckling

Axisx-x

Axisy-y

Thickness

t w

Width

b f

Thickness

t fk

d/Af

I xS x

r xI y

S yr y

(in2)

(in)

(in)

(in)

(in)

(in)

(in4)

(in3)

(in)

(in4)

(in3)

(in)

W30×1

32

38.9

30.31

0.615

10.545

113/4

2.874

5770

380

12.2

196

37.2

2.25

W30×1

24

36.5

30.17

0.585

10.515

0.93

111/16

3.085

5360

355

12.1

181

34.4

2.23

W30×1

16

34.2

30.01

0.565

10.495

0.85

15/8

3.364

4930

329

12

164

31.3

2.19

W30×1

08

31.7

29.83

0.545

10.475

0.76

19/16

3.747

4470

299

11.9

146

27.9

2.15

W30×9

929.1

29.65

0.52

10.45

0.67

17/16

4.235

3990

269

11.7

128

24.5

2.1

W30×9

026.4

29.53

0.47

10.4

0.61

15/16

4.655

3620

245

11.7

115

22.1

2.09

W27×5

39

158

32.52

1.97

15.255

3.54

41/4

0.602

25500

1570

12.7

2110

277

3.66

W27×4

48

131

31.42

1.65

14.94

2.99

311/16

0.703

20400

1300

12.5

1670

224

3.57

W27×3

68

108

30.39

1.38

14.665

2.48

33/16

0.836

16100

1060

12.2

1310

179

3.48

W27×3

07

90.2

29.61

1.16

14.445

2.09

213/16

0.981

13100

884

12

1050

146

3.42

W27×2

58

75.7

28.98

0.98

14.27

1.77

21/2

1.147

10800

742

11.9

859

120

3.37

W27×2

35

69.1

28.66

0.91

14.19

1.61

25/16

1.254

9660

674

11.8

768

108

3.33

W27×2

17

63.8

28.43

0.83

14.115

1.5

23/16

1.343

8870

624

11.8

704

99.8

3.32

W27×1

94

57

28.11

0.75

14.035

1.34

21/16

1.495

7820

556

11.7

618

88.1

3.29

W27×1

78

52.3

27.81

0.725

14.085

1.19

17/8

1.659

6990

502

11.6

555

78.8

3.26

376


W27×1

61

47.4

27.59

0.66

14.02

1.08

113/16

1.822

6280

455

11.5

497

70.9

3.24

W27×1

46

42.9

27.38

0.605

13.965

0.975

111/16

2.011

5630

411

11.4

443

63.5

3.21

W27×1

29

37.8

27.63

0.61

10.01

1.1

113/16

2.509

4760

345

11.2

184

36.8

2.21

W27×1

14

33.5

27.29

0.57

10.07

0.93

15/8

2.914

4090

299

11

159

31.5

2.18

W27×1

02

30

27.09

0.515

10.015

0.83

19/16

3.259

3620

267

11

139

27.8

2.15

W27×9

427.7

26.92

0.49

9.99

0.745

17/16

3.617

3270

243

10.9

124

24.8

2.12

W27×8

424.8

26.71

0.46

9.96

0.64

13/8

4.190

2850

213

10.7

106

21.2

2.07

W24×4

92

144

29.65

1.97

14.115

3.54

45/16

0.593

19100

1290

11.5

1670

237

3.41

W24×4

08

119

28.54

1.65

13.8

2.99

33/4

0.692

15100

1060

11.3

1320

191

3.33

W24×3

35

98.4

27.52

1.38

13.52

2.48

31/4

0.821

11900

864

11

1030

152

3.23

W24×2

79

82

26.73

1.16

13.305

2.09

27/8

0.961

9600

718

10.8

823

124

3.17

W24×2

50

73.5

26.34

1.04

13.185

1.89

211/16

1.057

8490

644

10.7

724

110

3.14

W24×2

29

67.2

26.02

0.96

13.11

1.73

21/2

1.147

7650

588

10.7

651

99.4

3.11

W24×2

07

60.7

25.71

0.87

13.01

1.57

23/8

1.259

6820

531

10.6

578

88.8

3.08

W24×1

92

56.3

25.47

0.81

12.95

1.46

21/4

1.347

6260

491

10.5

530

81.8

3.07

W24×1

76

51.7

25.24

0.75

12.89

1.34

21/8

1.461

5680

450

10.5

479

74.3

3.04

W24×1

62

47.7

25

0.705

12.955

1.22

21.582

5170

414

10.4

443

68.4

3.05

W24×1

46

43

24.74

0.65

12.9

1.09

17/8

1.759

4580

371

10.3

391

60.5

3.01

W24×1

31

38.5

24.48

0.605

12.855

0.96

13/4

1.984

4020

329

10.2

340

53

2.97

(continued)

377


Designation

Area

ADepth

d

Web

Flange

Lateral

Buckling

Axisx-x

Axisy-y

Thickness

t w

Width

b f

Thickness

t fk

d/Af

I xS x

r xI y

S yr y

(in2)

(in)

(in)

(in)

(in)

(in)

(in4)

(in3)

(in)

(in4)

(in3)

(in)

W24×1

17

34.4

24.26

0.55

12.8

0.85

15/8

2.230

3540

291

10.1

297

46.5

2.94

W24×1

04

30.6

24.06

0.5

12.75

0.75

11/2

2.516

3100

258

10.1

259

40.7

2.91

W24×1

03

30.3

24.53

0.55

90.98

13/4

2.781

3000

245

9.96

119

26.5

1.99

W24×9

427.7

24.31

0.515

9.065

0.875

15/8

3.065

2700

222

9.87

109

24

1.98

W24×8

424.7

24.1

0.47

9.02

0.77

19/16

3.470

2370

196

9.79

94.4

20.9

1.95

W24×7

622.4

23.92

0.44

8.99

0.68

17/16

3.913

2100

176

9.69

82.5

18.4

1.92

W24×6

820.1

23.73

0.415

8.965

0.585

13/8

4.525

1830

154

9.55

70.4

15.7

1.87

W24×6

218.2

23.74

0.43

7.04

0.59

13/8

5.716

1550

131

9.23

34.5

9.8

1.38

W24×5

516.2

23.57

0.395

7.005

0.505

15/16

6.663

1350

114

9.11

29.1

8.3

1.34

W21×2

01

59.2

23.03

0.91

12.575

1.63

23/8

1.124

5310

461

9.47

542

86.1

3.02

W21×1

82

53.6

22.72

0.83

12.5

1.48

21/4

1.228

4730

417

9.4

483

77.2

3

W21×1

66

48.8

22.48

0.75

12.42

1.36

21/8

1.331

4280

380

9.36

435

70.1

2.98

W21×1

47

43.2

22.06

0.72

12.51

1.15

17/8

1.533

3630

329

9.17

376

60.1

2.95

W21×1

32

38.8

21.83

0.65

12.44

1.035

113/16

1.695

3220

295

9.12

333

53.5

2.93

W21×1

22

35.9

21.68

0.6

12.39

0.96

111/16

1.823

2960

273

9.09

305

49.2

2.92

378


W21×1

11

32.7

21.51

0.55

12.34

0.875

15/8

1.992

2670

249

9.05

274

44.5

2.9

W21×1

01

29.8

21.36

0.5

12.29

0.8

19/16

2.172

2420

227

9.02

248

40.3

2.89

W21×9

327.3

21.62

0.58

8.42

0.93

111/16

2.761

2070

192

8.7

92.9

22.1

1.84

W21×8

324.3

21.43

0.515

8.355

0.835

19/16

3.072

1830

171

8.67

81.4

19.5

1.83

W21×7

321.5

21.24

0.455

8.295

0.74

11/2

3.460

1600

151

8.64

70.6

17

1.81

W21×6

820

21.13

0.43

8.27

0.685

17/16

3.730

1480

140

8.6

64.7

15.7

1.8

W21×6

218.3

20.99

0.4

8.24

0.615

13/8

4.142

1330

127

8.54

57.5

13.9

1.77

W21×5

716.7

21.06

0.405

6.555

0.65

13/8

4.943

1170

111

8.36

30.6

9.35

1.35

W21×5

014.7

20.83

0.38

6.53

0.535

15/16

5.962

984

94.5

8.18

24.9

7.64

1.3

W21×4

413

20.66

0.35

6.5

0.45

13/16

7.063

843

81.6

8.06

20.7

6.36

1.26

W18×3

11

91.5

22.32

1.52

12.005

2.74

37/16

0.679

6960

624

8.72

795

132

2.95

W18×2

83

83.2

21.85

1.4

11.89

2.5

33/16

0.735

6160

564

8.61

704

118

2.91

W18×2

58

75.9

21.46

1.28

11.77

2.3

30.793

5510

514

8.53

628

107

2.88

W18×2

34

68.8

21.06

1.16

11.65

2.11

23/4

0.857

4900

466

8.44

558

95.8

2.85

W18×2

11

62.1

20.67

1.06

11.555

1.91

29/16

0.937

4330

419

8.35

493

85.3

2.82

W18×1

92

56.4

20.35

0.96

11.455

1.75

27/16

1.015

3870

380

8.28

440

76.8

2.79

W18×1

75

51.3

20.04

0.89

11.375

1.59

21/4

1.108

3450

344

8.2

391

68.8

2.76

W18×1

58

46.3

19.72

0.81

11.3

1.44

21/8

1.212

3060

310

8.12

347

61.4

2.74

W18×1

43

42.1

19.49

0.73

11.22

1.32

21.316

2750

282

8.09

311

55.5

2.72

(continued)

379


Designation

Area

ADepth

d

Web

Flange

Lateral

Buckling

Axisx-x

Axisy-y

Thickness

t w

Width

b f

Thickness

t fk

d/Af

I xS x

r xI y

S yr y

(in2)

(in)

(in)

(in)

(in)

(in)

(in4)

(in3)

(in)

(in4)

(in3)

(in)

W18×1

30

38.2

19.25

0.67

11.16

1.2

17/8

1.437

2460

256

8.03

278

49.9

2.7

W18×1

19

35.1

18.97

0.655

11.265

1.06

13/4

1.589

2190

231

7.9

253

44.9

2.69

W18×1

06

31.1

18.73

0.59

11.2

0.94

15/8

1.779

1910

204

7.84

220

39.4

2.66

W18×9

728.5

18.59

0.535

11.145

0.87

19/16

1.917

1750

188

7.82

201

36.1

2.65

W18×8

625.3

18.39

0.48

11.09

0.77

17/16

2.154

1530

166

7.77

175

31.6

2.63

W18×7

622.3

18.21

0.425

11.035

0.68

13/8

2.427

1330

146

7.73

152

27.6

2.61

W18×7

120.8

18.47

0.495

7.635

0.81

11/2

2.987

1170

127

7.5

60.3

15.8

1.7

W18×6

519.1

18.35

0.45

7.59

0.75

17/16

3.224

1070

117

7.49

54.8

14.4

1.69

W18×6

017.6

18.24

0.415

7.555

0.695

13/8

3.474

984

108

7.47

50.1

13.3

1.69

W18×5

516.2

18.11

0.39

7.53

0.63

15/16

3.818

890

98.3

7.41

44.9

11.9

1.67

W18×5

014.7

17.99

0.355

7.495

0.57

11/4

4.211

800

88.9

7.38

40.1

10.7

1.65

W18×4

613.5

18.06

0.36

6.06

0.605

11/4

4.926

712

78.8

7.25

22.5

7.43

1.29

W18×4

011.8

17.9

0.315

6.015

0.525

13/16

5.668

612

68.4

7.21

19.1

6.35

1.27

W18×3

510.3

17.7

0.3

60.425

11/8

6.941

510

57.6

7.04

15.3

5.12

1.22

380


W16×1

00

29.4

16.97

0.585

10.425

0.985

111/16

1.653

1490

175

7.1

186

35.7

2.51

W16×8

926.2

16.75

0.525

10.365

0.875

19/16

1.847

1300

155

7.05

163

31.4

2.49

W16×7

722.6

16.52

0.455

10.295

0.76

17/16

2.111

1110

134

7138

26.9

2.47

W16×6

719.7

16.33

0.395

10.235

0.665

13/8

2.399

954

117

6.96

119

23.2

2.46

W16×5

716.8

16.43

0.43

7.12

0.715

13/8

3.227

758

92.2

6.72

43.1

12.1

1.6

W16×5

014.7

16.26

0.38

7.07

0.63

15/16

3.651

659

81

6.68

37.2

10.5

1.59

W16×4

513.3

16.13

0.345

7.035

0.565

11/4

4.058

586

72.7

6.65

32.8

9.34

1.57

W16×4

011.8

16.01

0.305

6.995

0.505

13/16

4.532

518

64.7

6.63

28.9

8.25

1.57

W16×3

610.6

15.86

0.295

6.985

0.43

11/8

5.280

448

56.5

6.51

24.5

71.52

W16×3

19.12

15.88

0.275

5.525

0.44

11/8

6.532

375

47.2

6.41

12.4

4.49

1.17

W16×2

67.68

15.69

0.25

5.5

0.345

11/16

8.269

301

38.4

6.26

9.59

3.49

1.12

W14×8

08

237

22.84

3.74

18.56

5.12

513/16

0.240

16000

1400

8.21

5510

594

4.82

W14×7

30

215

22.42

3.07

17.89

4.91

59/16

0.255

14300

1280

8.17

4720

527

4.69

W14×6

65

196

21.64

2.83

17.65

4.52

53/16

0.271

12400

1150

7.98

4170

472

4.62

W14×6

05

178

20.92

2.595

17.415

4.16

413/16

0.289

10800

1040

7.8

3680

423

4.55

W14×5

50

162

20.24

2.38

17.2

3.82

41/2

0.308

9430

931

7.63

3250

378

4.49

W14×5

00

147

19.6

2.19

17.01

3.5

43/16

0.329

8210

838

7.48

2880

339

4.43

W14×4

55

134

19.02

2.015

16.835

3.21

37/8

0.352

7190

756

7.33

2560

304

4.38

W14×4

26

125

18.67

1.875

16.695

3.035

311/16

0.368

6600

707

7.26

2360

283

4.34

(continued)

381


Designation

Area

ADepth

d

Web

Flange

Lateral

Buckling

Axisx-x

Axisy-y

Thickness

t w

Width

b f

Thickness

t fk

d/Af

I xS x

r xI y

S yr y

(in2)

(in)

(in)

(in)

(in)

(in)

(in4)

(in3)

(in)

(in4)

(in3)

(in)

W14×3

98

117

18.29

1.77

16.59

2.845

31/2

0.388

6000

656

7.16

2170

262

4.31

W14×3

70

109

17.92

1.655

16.475

2.66

35/16

0.409

5440

607

7.07

1990

241

4.27

W14×3

42

101

17.54

1.54

16.36

2.47

31/8

0.434

4900

559

6.98

1810

221

4.24

W14×3

11

91.4

17.12

1.41

16.23

2.26

215/16

0.467

4330

506

6.88

1610

199

4.2

W14×2

83

83.3

16.74

1.29

16.11

2.07

23/4

0.502

3840

459

6.79

1440

179

4.17

W14×2

57

75.6

16.38

1.175

15.995

1.89

29/16

0.542

3400

415

6.71

1290

161

4.13

W14×2

33

68.5

16.04

1.07

15.89

1.72

23/8

0.587

3010

375

6.63

1150

145

4.1

W14×2

11

62

15.72

0.98

15.8

1.56

21/4

0.638

2660

338

6.55

1030

130

4.07

W14×1

93

56.8

15.48

0.89

15.71

1.44

21/8

0.684

2400

310

6.5

931

119

4.05

W14×1

76

51.8

15.22

0.83

15.65

1.31

20.742

2140

281

6.43

838

107

4.02

W14×1

59

46.7

14.98

0.745

15.565

1.19

17/8

0.809

1900

254

6.38

748

96.2

4

W14×1

45

42.7

14.78

0.68

15.5

1.09

13/4

0.875

1710

232

6.33

677

87.3

3.98

W14×1

32

38.8

14.66

0.645

14.725

1.03

111/16

0.967

1530

209

6.28

548

74.5

3.76

W14×1

20

35.3

14.48

0.59

14.67

0.94

15/8

1.050

1380

190

6.24

495

67.5

3.74

W14×1

09

32

14.32

0.525

14.605

0.86

19/16

1.140

1240

173

6.22

447

61.2

3.73

W14×9

929.1

14.16

0.485

14.565

0.78

17/16

1.246

1110

157

6.17

402

55.2

3.71

382


W14×9

026.5

14.02

0.44

14.52

0.71

13/8

1.360

999

143

6.14

362

49.9

3.7

W14×8

224.1

14.31

0.51

10.13

0.855

15/8

1.652

882

123

6.05

148

29.3

2.48

W14×7

421.8

14.17

0.45

10.07

0.785

19/16

1.793

796

112

6.04

134

26.6

2.48

W14×6

820

14.04

0.415

10.035

0.72

11/2

1.943

723

103

6.01

121

24.2

2.46

W14×6

117.9

13.89

0.375

9.995

0.645

17/16

2.155

640

92.2

5.98

107

21.5

2.45

W14×5

315.6

13.92

0.37

8.06

0.66

17/16

2.617

541

77.8

5.89

57.7

14.3

1.92

W14×4

814.1

13.79

0.34

8.03

0.595

13/8

2.886

485

70.3

5.85

51.4

12.8

1.91

W14×4

312.6

13.66

0.305

7.995

0.53

15/16

3.224

428

62.7

5.82

45.2

11.3

1.89

W14×3

811.2

14.1

0.31

6.77

0.515

11/16

4.044

385

54.6

5.87

26.7

7.88

1.55

W14×3

410

13.98

0.285

6.745

0.455

14.555

340

48.6

5.83

23.3

6.91

1.53

W14×3

08.85

13.84

0.27

6.73

0.385

15/16

5.341

291

42

5.73

19.6

5.82

1.49

W14×2

67.69

13.91

0.255

5.025

0.42

15/16

6.591

245

35.3

5.65

8.91

3.54

1.08

W14×2

26.49

13.74

0.23

50.335

7/8

8.203

199

29

5.54

72.8

1.04

W12×3

36

98.8

16.82

1.775

13.385

2.955

311/16

0.425

4060

483

6.41

1190

177

3.47

W12×3

05

89.6

16.32

1.625

13.235

2.705

37/16

0.456

3550

435

6.29

1050

159

3.42

W12×2

79

81.9

15.85

1.53

13.14

2.47

33/16

0.488

3110

393

6.16

937

143

3.38

W12×2

52

74.1

15.41

1.395

13.005

2.25

215/16

0.527

2720

353

6.06

828

127

3.34

W12×2

30

67.7

15.05

1.285

12.895

2.07

23/4

0.564

2420

321

5.97

742

115

3.31

W12×2

10

61.8

14.71

1.18

12.79

1.9

25/8

0.605

2140

292

5.89

664

104

3.28

(continued)

383


Designation

Area

ADepth

d

Web

Flange

Lateral

Buckling

Axisx-x

Axisy-y

Thickness

t w

Width

b f

Thickness

t fk

d/Af

I xS x

r xI y

S yr y

(in2)

(in)

(in)

(in)

(in)

(in)

(in4)

(in3)

(in)

(in4)

(in3)

(in)

W12×1

90

55.8

14.38

1.06

12.67

1.735

27/16

0.654

1890

263

5.82

589

93

3.25

W12×1

70

50

14.03

0.96

12.57

1.56

21/4

0.715

1650

235

5.74

517

82.3

3.22

W12×1

52

44.7

13.71

0.87

12.48

1.4

21/8

0.785

1430

209

5.66

454

72.8

3.19

W12×1

36

39.9

13.41

0.79

12.4

1.25

115/16

0.865

1240

186

5.58

398

64.2

3.16

W12×1

20

35.3

13.12

0.71

12.32

1.105

113/16

0.964

1070

163

5.51

345

56

3.13

W12×1

06

31.2

12.89

0.61

12.22

0.99

111/16

1.065

933

145

5.47

301

49.3

3.11

W12×9

628.2

12.71

0.55

12.16

0.9

15/8

1.161

833

131

5.44

270

44.4

3.09

W12×8

725.6

12.53

0.515

12.125

0.81

11/2

1.276

740

118

5.38

241

39.7

3.07

W12×7

923.2

12.38

0.47

12.08

0.735

17/16

1.394

662

107

5.34

216

35.8

3.05

W12×7

221.1

12.25

0.43

12.04

0.67

13/8

1.519

597

97.4

5.31

195

32.4

3.04

W12×6

519.1

12.12

0.39

12

0.605

15/16

1.669

533

87.9

5.28

174

29.1

3.02

W12×5

817

12.19

0.36

10.01

0.64

13/8

1.903

475

78

5.28

107

21.4

2.51

W12×5

315.6

12.06

0.345

9.995

0.575

11/4

2.098

425

70.6

5.23

95.8

19.2

2.48

W12×5

014.7

12.19

0.37

8.08

0.64

13/8

2.357

394

64.7

5.18

56.3

13.9

1.96

W12×4

513.2

12.06

0.335

8.045

0.575

11/4

2.607

350

58.1

5.15

50

12.4

1.94

W12×4

011.8

11.94

0.295

8.005

0.515

11/4

2.896

310

51.9

5.13

44.1

11

1.93

384


W12×3

510.3

12.5

0.3

6.56

0.52

13.664

285

45.6

5.25

24.5

7.47

1.54

W12×3

08.79

12.34

0.26

6.52

0.44

15/16

4.301

238

38.6

5.21

20.3

6.24

1.52

W12×2

67.65

12.22

0.23

6.49

0.38

7/8

4.955

204

33.4

5.17

17.3

5.34

1.51

W12×2

26.48

12.31

0.26

4.03

0.425

7/8

7.187

156

25.4

4.91

4.66

2.31

0.847

W12×1

95.57

12.16

0.235

4.005

0.35

13/16

8.675

130

21.3

4.82

3.76

1.88

0.822

W12×1

64.71

11.99

0.22

3.99

0.265

3/4

11.340

103

17.1

4.67

2.82

1.41

0.773

W12×1

44.16

11.91

0.2

3.97

0.225

11/16

13.333

88.6

14.9

4.62

2.36

1.19

0.753

W10×1

12

32.9

11.36

0.755

10.415

1.25

17/8

0.873

716

126

4.66

236

45.3

2.68

W10×1

00

29.4

11.1

0.68

10.34

1.12

13/4

0.958

623

112

4.6

207

40

2.65

W10×8

825.9

10.84

0.605

10.265

0.99

15/8

1.067

534

98.5

4.54

179

34.8

2.63

W10×7

722.6

10.6

0.53

10.19

0.87

11/2

1.196

455

85.9

4.49

154

30.1

2.6

W10×6

820

10.4

0.47

10.13

0.77

13/8

1.333

394

75.7

4.44

134

26.4

2.59

W10×6

017.6

10.22

0.42

10.08

0.68

15/16

1.491

341

66.7

4.39

116

23

2.57

W10×5

415.8

10.09

0.37

10.03

0.615

11/4

1.636

303

60

4.37

103

20.6

2.56

W10×4

914.4

9.98

0.34

10

0.56

13/16

1.782

272

54.6

4.35

93.4

18.7

2.54

W10×4

513.3

10.1

0.35

8.02

0.62

11/4

2.031

248

49.1

4.32

53.4

13.3

2.01

W10×3

911.5

9.92

0.315

7.985

0.53

11/8

2.344

209

42.1

4.27

45

11.3

1.98

W10×3

39.71

9.73

0.29

7.96

0.435

11/16

2.810

170

35

4.19

36.6

9.2

1.94

W10×3

08.84

10.47

0.3

5.81

0.51

15/16

3.533

170

32.4

4.38

16.7

5.75

1.37

(continued)

385


Designation

Area

ADepth

d

Web

Flange

Lateral

Buckling

Axisx-x

Axisy-y

Thickness

t w

Width

b f

Thickness

t fk

d/Af

I xS x

r xI y

S yr y

(in2)

(in)

(in)

(in)

(in)

(in)

(in4)

(in3)

(in)

(in4)

(in3)

(in)

W10×2

67.61

10.33

0.26

5.77

0.44

7/8

4.069

144

27.9

4.35

14.1

4.89

1.36

W10×2

26.49

10.17

0.24

5.75

0.36

3/4

4.913

118

23.2

4.27

11.4

3.97

1.33

W10×1

95.62

10.24

0.25

4.02

0.395

13/16

6.449

96.3

18.8

4.14

4.29

2.14

0.874

W10×1

74.99

10.11

0.24

4.01

0.33

3/4

7.640

81.9

16.2

4.05

3.56

1.78

0.844

W10×1

54.41

9.99

0.23

40.27

11/16

9.250

68.9

13.8

3.95

2.89

1.45

0.81

W10×1

23.54

9.87

0.19

3.96

0.21

5/8

11.869

53.8

10.9

3.9

2.18

1.1

0.785

W8×6

719.7

90.57

8.28

0.935

17/16

1.163

272

60.4

3.72

88.6

21.4

2.12

W8×5

817.1

8.75

0.51

8.22

0.81

15/16

1.314

228

52

3.65

75.1

18.3

2.1

W8×4

814.1

8.5

0.4

8.11

0.685

13/16

1.530

184

43.3

3.61

60.9

15

2.08

W8×4

011.7

8.25

0.36

8.07

0.56

11/16

1.826

146

35.5

3.53

49.1

12.2

2.04

W8×3

510.3

8.12

0.31

8.02

0.495

12.045

127

31.2

3.51

42.6

10.6

2.03

W8×3

19.13

80.285

7.995

0.435

15/16

2.300

110

27.5

3.47

37.1

9.27

2.02

W8×2

88.25

8.06

0.285

6.535

0.465

15/16

2.652

98

24.3

3.45

21.7

6.63

1.62

W8×2

47.08

7.93

0.245

6.495

0.4

7/8

3.052

82.8

20.9

3.42

18.3

5.63

1.61

W8×2

16.16

8.28

0.25

5.27

0.4

13/16

3.928

75.3

18.2

3.49

9.77

3.71

1.26

386


W8×1

85.26

8.14

0.23

5.25

0.33

3/4

4.698

61.9

15.2

3.43

7.97

3.04

1.23

W8×1

54.44

8.11

0.245

4.015

0.315

3/4

6.412

48

11.8

3.29

3.41

1.7

0.876

W8×1

33.84

7.99

0.23

40.255

11/16

7.833

39.6

9.91

3.21

2.73

1.37

0.843

W8×1

02.96

7.89

0.17

3.94

0.205

5/8

9.768

30.8

7.81

3.22

2.09

1.06

0.841

W6×2

57.34

6.38

0.32

6.08

0.455

13/16

2.306

53.4

16.7

2.7

17.1

5.61

1.52

W6×2

05.87

6.2

0.26

6.02

0.365

3/4

2.822

41.4

13.4

2.66

13.3

4.41

1.5

W6×1

54.43

5.99

0.23

5.99

0.26

5/8

3.846

29.1

9.72

2.56

9.32

3.11

1.46

W6×1

64.74

6.28

0.26

4.03

0.405

3/4

3.848

32.1

10.2

2.6

4.43

2.2

0.966

W6×1

23.55

6.03

0.23

40.28

5/8

5.384

22.1

7.31

2.49

2.99

1.5

0.918

W6×9

2.68

5.9

0.17

3.94

0.215

9/16

6.965

16.4

5.56

2.47

2.19

1.11

0.905

W5×1

95.54

5.15

0.27

5.03

0.43

13/16

2.381

26.2

10.2

2.17

9.13

3.63

1.28

W5×1

64.68

5.01

0.24

50.36

3/4

2.783

21.3

8.51

2.13

7.51

31.27

W4×1

33.83

4.16

0.28

4.06

0.345

11/16

2.970

11.3

5.46

1.72

3.86

1.9

1

387


American

Stan

dard

C-C

hann

el

Designation

Area

ADepth

d

Web

Flange

Lateral

Buckling

Axisx-x

Axisy-y

Thickness

t w

Width

b f

Thickness

t fk

d/Af

I xS x

r xI y

S yr y

(in2)

(in)

(in)

(in)

(in)

(in)

(in4)

(in3)

(in)

(in4)

(in3)

(in)

C15×5

014.7

15.0

0.716

3.72

0.650

1.44

6.203

404

53.8

5.24

11.0

3.77

0.865

C15×4

011.8

15.0

0.520

3.52

0.650

1.44

6.556

348

46.5

5.43

9.17

3.34

0.883

C15×3

3.9

10.0

15.0

0.400

3.40

0.650

1.44

6.787

315

42.0

5.61

8.07

3.09

0.901

C12×3

08.81

12.0

0.510

3.17

0.501

1.13

7.556

162

27.0

4.29

5.12

2.05

0.762

C12×2

57.34

12.0

0.387

3.05

0.501

1.13

7.853

144

24.0

4.43

4.45

1.87

0.779

C12×2

0.7

6.08

12.0

0.282

2.94

0.501

1.13

8.147

129

21.5

4.61

3.86

1.72

0.797

C10×3

08.81

10.0

0.673

3.03

0.436

1.00

7.570

103

20.7

3.43

3.93

1.65

0.668

C10×2

57.35

10.0

0.526

2.89

0.436

1.00

7.936

91.1

18.2

3.52

3.34

1.47

0.675

C10×2

05.87

10.0

0.379

2.74

0.436

1.00

8.371

78.9

15.8

3.67

2.80

1.31

0.690

C10×1

5.3

4.48

10.0

0.240

2.60

0.436

1.00

8.821

67.3

13.5

3.88

2.27

1.15

0.711

388


C9×2

05.87

9.00

0.448

2.65

0.413

1.00

8.223

60.9

13.5

3.22

2.41

1.17

0.640

C9×1

54.40

9.00

0.285

2.49

0.413

1.00

8.752

51.0

11.3

3.40

1.91

1.01

0.659

C9×1

3.4

3.94

9.00

0.233

2.43

0.413

1.00

8.968

47.8

10.6

3.48

1.75

0.954

0.666

C8×1

8.75

5.51

8.00

0.487

2.53

0.390

0.938

8.108

43.9

11.0

2.82

1.97

1.01

0.598

C8×1

3.75

4.03

8.00

0.303

2.34

0.390

0.938

8.766

36.1

9.02

2.99

1.52

0.848

0.613

C8×1

1.5

3.37

8.00

0.220

2.26

0.390

0.938

9.076

32.5

8.14

3.11

1.31

0.775

0.623

C7×1

4.75

4.33

7.00

0.419

2.30

0.366

0.875

8.316

27.2

7.78

2.51

1.37

0.772

0.561

C7×1

2.25

3.59

7.00

0.314

2.19

0.366

0.875

8.733

24.2

6.92

2.59

1.16

0.696

0.568

C7×9

.82.87

7.00

0.210

2.09

0.366

0.875

9.151

21.2

6.07

2.72

0.957

0.617

0.578

C6×1

33.82

6.00

0.437

2.16

0.343

0.813

8.098

17.3

5.78

2.13

1.05

0.638

0.524

C6×1

0.5

3.07

6.00

0.314

2.03

0.343

0.813

8.617

15.1

5.04

2.22

0.860

0.561

0.529

C6×8

.22.39

6.00

0.200

1.92

0.343

0.813

9.111

13.1

4.35

2.34

0.687

0.488

0.536

(continued)

389


Designation

Area

ADepth

d

Web

Flange

Lateral

Buckling

Axisx-x

Axisy-y

Thickness

t w

Width

b f

Thickness

t fk

d/Af

I xS x

r xI y

S yr y

(in2)

(in)

(in)

(in)

(in)

(in)

(in4)

(in3)

(in)

(in4)

(in3)

(in)

C5×9

2.64

5.00

0.325

1.89

0.320

0.750

8.267

8.89

3.56

1.84

0.624

0.444

0.486

C5×6

.71.97

5.00

0.190

1.75

0.320

0.750

8.929

7.48

2.99

1.95

0.470

0.372

0.489

C4×7

.25

2.13

4.00

0.321

1.72

0.296

0.750

7.857

4.58

2.29

1.47

0.425

0.337

0.447

C4×6

.25

1.77

4.00

0.247

1.65

0.272

0.750

8.913

4.00

2.00

1.50

0.345

0.284

0.441

C4×5

.41.58

4.00

0.184

1.58

0.296

0.750

8.553

3.85

1.92

1.56

0.312

0.277

0.444

C4×4

.51.38

4.00

0.125

1.58

0.296

0.750

8.553

3.65

1.83

1.63

0.289

0.265

0.457

C3×6

1.76

3.00

0.356

1.60

0.273

0.688

6.868

2.07

1.38

1.09

0.300

0.263

0.413

C3×5

1.47

3.00

0.258

1.50

0.273

0.688

7.326

1.85

1.23

1.12

0.241

0.228

0.405

C3×4

.11.20

3.00

0.170

1.41

0.273

0.688

7.794

1.65

1.10

1.18

0.191

0.196

0.398

C3×3

.51.09

3.00

0.132

1.37

0.273

0.688

8.021

1.57

1.04

1.20

0.169

0.182

0.394

390


APPENDIX 2

STEEL PIPE

PIPE Dimensions and Properties

Dimensions Properties

Weight

lb per ft

Schedule

No.Nominal

Diameter

in.

Outside

Diameter

in.

Inside

Diameter

in.

Wall

Thickness

in.

Ain.2

Iin.4

Sin.3

rin.

Standard Weight

1/2 0.840 0.622 0.109 0.85 0.250 0.017 0.041 0.261 40

3/4 1.050 0.824 0.113 1.13 0.333 0.037 0.071 0.334 40

1 1.315 1.049 0.133 1.68 0.494 0.087 0.133 0.421 40

1-1/4 1.660 1.380 0.140 2.27 0.669 0.195 0.235 0.540 40

1-1/2 1.900 1.610 0.145 2.72 0.799 0.310 0.326 0.623 40

2 2.375 2.067 0.154 3.65 1.075 0.666 0.561 0.787 40

2-1/2 2.875 2.469 0.203 5.79 1.704 1.530 1.064 0.947 40

3 3.500 3.068 0.216 7.58 2.228 3.017 1.724 1.164 40

3-1/2 4.000 3.548 0.226 9.11 2.680 4.788 2.394 1.337 40

4 4.500 4.026 0.237 10.79 3.174 7.233 3.214 1.510 40

5 5.563 5.047 0.258 14.62 4.300 15.16 5.451 1.878 40

6 6.625 6.065 0.280 18.97 5.581 28.14 8.496 2.245 40

8 8.625 7.981 0.322 28.55 8.399 72.49 16.81 2.938 40

10 10.75 10.02 0.365 40.48 11.91 160.7 29.90 3.674 40

12 12.75 12.00 0.375 49.56 14.58 279.3 43.82 4.377 —

391


392 STEEL PIPE

PIPE Dimensions and Properties

Dimensions Properties

Weight

lb per ft

Schedule

No.Nominal

Diameter

in.

Outside

Diameter

in.

Inside

Diameter

in.

Wall

Thickness

in.

Ain.2

Iin.4

Sin.3

rin.

Extra Strong

1/2 0.840 0.546 0.147 1.090 0.320 0.020 0.048 0.250 80

3/4 1.050 0.742 0.154 1.470 0.433 0.045 0.085 0.321 80

1 1.315 0.957 0.179 2.170 0.639 0.106 0.161 0.407 80

1-1/4 1.660 1.278 0.191 3.000 0.881 0.242 0.291 0.524 80

1-1/2 1.900 1.500 0.200 3.630 1.068 0.391 0.412 0.605 80

2 2.375 1.939 0.218 5.020 1.477 0.868 0.731 0.766 80

2-1/2 2.875 2.323 0.276 7.660 2.254 1.924 1.339 0.924 80

3 3.500 2.900 0.300 10.25 3.016 3.894 2.225 1.136 80

3-1/2 4.000 3.364 0.318 12.50 3.678 6.280 3.140 1.307 80

4 4.500 3.826 0.337 14.98 4.407 9.610 4.271 1.477 80

5 5.563 4.813 0.375 20.78 6.112 20.67 7.431 1.839 80

6 6.625 5.761 0.432 28.57 8.405 40.49 12.22 2.195 80

8 8.625 7.625 0.500 43.39 12.76 105.7 24.51 2.878 80

10 10.75 9.750 0.500 54.74 16.10 212.0 39.43 3.628 60

12 12.75 11.75 0.500 65.42 19.24 361.5 56.71 4.335 —

Double-Extra Strong

2 2.375 1.503 0.436 9.03 2.66 1.31 1.10 0.703 —

2-1/2 2.875 1.771 0.552 13.69 4.03 2.87 2.00 0.844 —

3 3.500 2.300 0.600 18.58 5.47 5.99 3.42 1.047 —

4 4.500 3.152 0.674 27.54 8.10 15.3 6.79 1.374 —

5 5.563 4.063 0.750 38.55 11.34 33.6 12.09 1.722 —

6 6.625 4.897 0.864 53.16 15.64 66.3 20.02 2.060 —

8 8.625 6.875 0.875 72.42 21.30 162 37.56 2.757 —


APPENDIX 3

H PILE (AISC)

393


HPSh

apes

Web

Flange

Lateral

Buckling

Axisx-x

Axisy-y

Designation

Area

ADepth

dThickness

t w

Width

b f

Thickness

t fk

d/Af

I xS x

r xI y

S yr y

(in2)

(in)

(in)

(in)

(in)

(in)

(in4)

(in3)

(in)

(in4)

(in3)

(in)

HP14X117

34.4

14.21

0.805

14.885

0.805

11/16

1.186

1220

172

5.96

443

59.5

3.59

HP14X102

30

14.01

0.705

14.785

0.705

11.344

1050

150

5.92

380

51.4

3.56

HP14X89

26.1

13.83

0.615

14.695

0.615

15/16

1.530

904

131

5.88

326

44.3

3.53

HP14X73

21.4

13.61

0.505

14.585

0.505

7/8

1.848

729

107

5.84

261

35.8

3.49

HP13X100

29.4

13.15

0.765

13.205

0.765

17/16

1.302

886

135

5.49

294

44.5

3.16

HP13X87

25.5

12.95

0.665

13.105

0.665

13/8

1.486

755

117

5.45

250

38.1

3.13

HP13X73

21.6

12.75

0.565

13.005

0.565

11/4

1.735

630

98.8

5.4

207

31.9

3.1

HP13X60

17.5

12.54

0.46

12.9

0.46

11/8

2.113

503

80.3

5.36

165

25.5

3.07

HP12X84

24.6

12.28

0.685

12.295

0.685

13/8

1.458

650

106

5.14

213

34.6

2.94

HP12X85

21.8

12.13

0.605

12.215

0.605

15/16

1.641

569

92.8

5.11

186

30.4

2.92

HP12X86

18.4

11.94

0.515

12.125

0.515

11/4

1.912

472

79.1

5.06

153

25.3

2.88

HP12X87

15.5

11.78

0.435

12.045

0.435

11/8

2.248

393

66.8

5.03

127

21.1

2.86

HP10X57

16.8

9.99

0.656

10.225

0.656

13/16

1.489

294

58.8

4.18

101

19.7

2.45

HP10X42

12.4

9.7

0.415

10.075

0.415

11/16

2.320

210

43.4

4.13

71.7

14.2

2.41

HP8X36

10.6

8.02

0.445

8.155

0.445

15/16

2.210

119

29.8

3.36

40.3

9.88

1.95

394


APPENDIX 4

ALLOWABLE BUCKLING STRESS

AISC Allowable Compressive (Buckling) Stresses for A36 (36 KSI) Steel (Fbs)

kL/r Fbs (ksi) kL/r Fbs (ksi) kL/r Fbs (ksi) kL/r Fbs (ksi) kL/r Fbs (ksi)

1 21.56 41 19.11 81 15.24 121 10.14 161 5.76

2 21.52 42 19.02 82 15.14 122 9.98 162 5.69

3 21.48 43 18.93 83 15.02 123 9.86 163 5.62

4 21.44 44 18.84 84 14.91 124 9.71 164 5.55

5 21.39 45 18.78 85 14.78 125 9.56 165 5.48

6 21.35 46 18.69 86 14.67 126 9.41 166 5.41

7 21.31 47 18.6 87 14.56 127 9.26 167 5.34

8 21.25 48 18.51 88 14.44 128 9.12 168 5.27

9 21.21 49 18.42 89 14.32 129 8.98 169 5.23

10 21.16 50 18.35 90 14.21 130 8.85 170 5.16

11 21.11 51 18.26 91 14.09 131 8.71 171 5.09

12 21.06 52 18.17 92 13.97 132 8.58 172 5.05

13 21.01 53 18.08 93 13.84 133 8.45 173 4.99

14 20.96 54 17.99 94 13.72 134 8.32 174 4.93

15 20.89 55 17.91 95 13.6 135 8.21 175 4.87

16 20.84 56 17.82 96 13.48 136 8.08 176 4.81

395


396 ALLOWABLE BUCKLING STRESS

kL/r Fbs (ksi) kL/r Fbs (ksi) kL/r Fbs (ksi) kL/r Fbs (ksi) kL/r Fbs (ksi)

17 20.79 57 17.73 97 13.35 137 7.97 177 4.77

18 20.73 58 17.64 98 13.23 138 7.85 178 4.72

19 20.67 59 17.55 99 13.1 139 7.74 179 4.67

20 20.61 60 17.43 100 12.98 140 7.63 180 4.62

21 20.54 61 17.34 101 12.85 141 7.52 181 4.57

22 20.48 62 17.25 102 12.72 142 7.42 182 4.52

23 20.42 63 17.16 103 12.58 143 7.31 183 4.47

24 20.36 64 17.07 104 12.48 144 7.21 184 4.42

25 20.29 65 16.94 105 12.34 145 7.11 185 4.38

26 20.23 66 16.85 106 12.21 146 7.02 186 4.33

27 20.16 67 16.74 107 12.08 147 6.92 187 4.28

28 20.08 68 16.64 108 11.94 148 6.83 188 4.24

29 19.95 69 16.53 109 11.81 149 6.75 189 4.19

30 19.88 70 16.43 110 11.67 150 6.65 190 4.15

31 19.82 71 16.33 111 11.54 151 6.56 191 4.11

32 19.75 72 16.22 112 11.41 152 6.47 192 4.06

33 19.68 73 16.12 113 11.27 153 6.39 193 4.02

34 19.61 74 16.01 114 11.14 154 6.31 194 3.98

35 19.58 75 15.91 115 10.99 155 6.23 195 3.94

36 19.51 76 15.79 116 10.86 156 6.14 196 3.89

37 19.42 77 15.69 117 10.72 157 6.06 197 3.86

38 19.35 78 15.58 118 10.57 158 5.99 198 3.81

39 19.28 79 15.47 119 10.42 159 5.92 199 3.78

40 19.19 80 15.35 120 10.27 160 5.85 200 3.74


APPENDIX 5

SHEET PILE (SKYLINE)

397


Section

Thickness

Cross-

Sectional

Area

Weight

Section

Modulus

Momentof

Inertia

CoationArea

Width

Height

Flange

Web

Pile

Wall

Elastic

Both

Sides

WallSurface

(w)

(h)

(tf)

(tw)

inin

inin

in2/ft

lb/ft

lb/ft2

in3/ft

in4/ft

ft2/ftofsingle

ft2/ft2

(mm)

(mm)

(mm)

(mm)

(cm

2/m

)(kg/m

)(kg/m

2)

(cm

3/m

)(cm

4/m

)(m

2/m

)(m

2/m

2)

AZ12-700

27.56

12.36

0.335

0.335

5.82

45.49

19.81

22.4

138.3

5.61

1.22

700

314

8.5

8.5

123.2

67.7

96.7

1205

18880

1.71

1.22

AZ13-700

27.56

12.4

0.375

0.375

6.36

49.72

21.65

24.3

150.4

5.61

1.22

700

315

9.5

9.5

134.7

74

105.7

1305

20540

1.71

1.22

AZ13-700-10/10

27.56

12.42

0.394

0.394

6.63

51.85

22.58

25.2

156.5

5.61

1.22

700

316

10

10

140.4

77.2

110.2

1355

21370

1.71

1.22

AZ14-700

27.56

12.44

0.413

0.413

6.9

53.96

23.5

26.1

162.5

5.61

1.22

700

316

10.5

10.5

146.1

80.3

114.7

1405

22190

1.71

1.22

AZ12-770

30.31

13.52

0.335

0.335

5.67

48.78

19.31

23.2

156.9

6.10

1.20

770

343.5

8.5

8.5

120.1

72.6

94.3

1245

21430

1.86

1.20

AZ13-770

30.31

13.54

0.354

0.354

5.94

51.14

20.24

24.2

163.7

6.10

1.20

770

344

99

125.8

76.1

98.8

1300

22360

1.86

1.20

AZ14-770

30.31

13.56

0.375

0.375

6.21

53.42

21.14

25.2

170.6

6.10

1.20

770

344.5

9.5

9.5

131.5

79.5

103.2

1355

23300

1.86

1.2

AZ14-770-10/10

30.31

13.58

0.394

0.394

6.48

55.71

22.06

26.1

177.5

6.07

1.2

770

345

10

10

137.2

82.9

107.7

1405

24240

1.85

1.2

398


AZ18

24.8

14.96

0.375

0.375

7.11

49.99

24.19

33.5

250.4

5.64

1.35

630

380

9.5

9.5

150.4

74.4

118.1

1800

34200

1.72

1.35

AZ17-700

27.56

16.52

0.335

0.335

6.28

49.12

21.38

32.2

265.3

6.1

1.33

700

419.5

8.5

8.5

133

73.1

104.4

1730

36230

1.86

1.33

AZ18-700

27.56

16.54

0.354

0.354

6.58

51.41

22.39

33.5

276.8

6.1

1.33

700

420

99

139.2

76.5

109.3

1800

37800

1.86

1.33

AZ19-700

27.56

16.56

0.375

0.375

6.88

53.76

23.41

34.8

288.4

6.1

1.33

700

420.5

9.5

9.5

145.6

80

114.3

1870

39380

1.86

1.33

AZ20-700

27.56

16.58

0.394

0.394

7.18

56.11

24.43

36.2

299.9

6.1

1.33

700

421

10

10

152

83.5

119.3

1945

40960

1.86

1.33

AZ26

24.8

16.81

0.512

0.48

9.35

65.72

31.79

48.4

406.5

5.91

1.41

630

427

13

12.2

198

97.8

155.2

2600

55510

1.8

1.41

AZ24-700

27.56

18.07

0.441

0.441

8.23

64.3

28

45.2

408.8

6.33

1.38

700

459

11.2

11.2

174.1

95.7

136.7

2430

55820

1.93

1.38

AZ26-700

27.56

18.11

0.48

0.48

8.84

69.12

30.1

48.4

437.3

6.33

1.38

700

460

12.2

12.2

187.2

102.9

146.9

2600

59720

1.93

1.38

AZ28-700

27.56

18.15

0.52

0.52

9.46

73.93

32.19

51.3

465.9

6.33

1.38

700

461

13.2

13.2

200.2

110

157.2

2760

63620

1.93

1.38

AZ24-700N

27.56

18.07

0.492

0.354

7.71

60.28

26.26

45.3

409.3

6.3

1.37

700

459

12.5

9163.3

89.7

128.2

2435

55890

1.92

1.37

AZ26-700N

27.56

18.11

0.531

0.394

8.33

65.11

28.37

48.4

437.8

6.3

1.37

700

460

13.5

10

176.4

96.9

138.5

2600

59790

1.92

1.37

399


Section

Thickness

Cross-

Sectional

Area

Weight

Section

Modulus

Momentof

Inertia

CoationArea

Width

Height

Flange

Web

Pile

Wall

Elastic

Both

Sides

WallSurface

(w)

(h)

(tf)

(tw)

inin

inin

in2/ft

lb/ft

lb/ft2

in3/ft

in4/ft

ft2/ftofsingle

ft2/ft2

(mm)

(mm)

(mm)

(mm)

(cm

2/m

)(kg/m

)(kg/m

2)

(cm

3/m

)(cm

4/m

)(m

2/m

)(m

2/m

2)

AZ28-700N

27.56

18.15

0.571

0.433

8.95

69.95

30.46

51.4

466.5

6.3

1.37

700

461

14.5

11

189.5

104.1

148.7

2765

63700

1.92

1.37

AZ36-700N

27.56

19.65

0.591

0.441

10.2

79.7

34.61

66.8

656.2

6.76

1.47

700

499

15

11.2

216

118.6

169

3590

89610

2.06

1.47

AZ38-700N

27.56

19.69

0.63

0.48

10.87

84.94

37.07

70.6

694.5

6.76

1.47

700

500

16

12.2

230

126.4

181

3795

94840

2.06

1.47

AZ40-700N

27.56

19.72

0.669

0.52

11.53

90.18

39.32

74.3

732.9

6.76

1.47

700

501

17

13.2

244

134.2

192

3995

100080

2.06

1.47

AZ42-700N

27.56

19.65

0.709

0.551

12.22

95.49

41.57

78.2

766

6.76

1.47

700

499

18

14

259

142.1

203

4205

104930

2.06

1.47

AZ44-700N

27.56

19.69

0.748

0.591

12.89

100.73

43.83

81.9

804.1

6.76

1.47

700

500

19

15

273

149.9

214

4405

110150

2.06

1.47

AZ46-700N

27.56

19.72

0.787

0.63

13.55

105.97

46.08

85.7

842.2

6.76

1.47

700

501

20

16

287

157.7

225

4605

115370

2.06

1.47

AZ46

22.83

18.94

0.709

0.551

13.76

89.1

46.82

85.5

808.8

6.23

1.63

580

481

18

14

291.2

132.6

228.6

4595

110450

1.9

1.63

AZ48

22.83

18.98

0.748

0.591

14.48

93.81

49.28

89.3

847.1

6.23

1.63

580

482

19

15

306.5

139.6

240.6

4800

115670

1.9

1.63

AZ50

22.83

19.02

0.787

0.63

15.22

98.58

51.8

93.3

886.5

6.23

1.63

580

483

20

16

322.2

146.7

252.9

5015

121060

1.9

1.63

400


APPENDIX 6

WOOD PROPERTIES

Rectangular (Wood) PropertiesNominal

Size b × dStandard

Dressed Size

(S4S) b ×d inches× inches

Area of

Section

A in2

x-x AXIS (Strong) y-y AXIS (Weak) Approximate Weight

in Pound per Linear

Foot (lb/ft) of Piece

When Density

Equals 35 lb/ft3

Section

Modulus

SXX in3

Moment

of Inertia

IXX in4

Section

Modulus

SYY in3

Moment

of Inertia

IYY in4

1 × 3 3/4 × 2-1/2 1.875 0.781 0.977 0.234 0.088 0.456

1 × 4 3/4 × 3-1/2 2.625 1.531 2.680 0.328 0.123 0.638

1 × 6 3/4 × 5-1/2 4.125 3.781 10.40 0.516 0.193 1.003

1 × 8 3/4 × 7-1/4 5.438 6.571 23.82 0.680 0.255 1.322

1 × 10 3/4 × 9-1/4 6.938 10.70 49.47 0.867 0.325 1.686

1 × 12 3/4 × 11-1/4 8.438 15.82 88.99 1.055 0.396 2.051

2 × 3 1-1/2 × 2-1/2 3.750 1.563 1.953 0.938 0.703 0.911

2 × 4 1-1/2 × 3-1/2 5.250 3.063 5.359 1.313 0.984 1.276

2 × 5 1-1/2 × 4-1/2 6.750 5.063 11.39 1.688 1.266 1.641

2 × 6 1-1/2 × 5-1/2 8.250 7.563 20.80 2.063 1.547 2.005

2 × 8 1-1/2 × 7-1/4 10.88 13.15 47.66 2.720 2.040 2.644

2 × 10 1-1/2 × 9-1/4 13.88 21.40 98.97 3.470 2.603 3.374

2 × 12 1-1/2 × 11-1/4 16.88 31.65 178.0 4.220 3.165 4.103

2 × 14 1-1/2 × 13-1/4 19.88 43.90 290.8 4.970 3.728 4.832

401


402 WOOD PROPERTIES

Nominal

Size b × dStandard

Dressed Size


Area of

Section

A in2


in Pound per Linear


When Density

Equals 35 lb/ft3

Section

Modulus

SXX in3

Moment

of Inertia

IXX in4

Section

Modulus

SYY in3

Moment

of Inertia

IYY in4

3 × 4 2-1/2 × 3-1/2 8.750 5.104 8.932 3.646 4.557 2.127

3 × 5 2-1/2 × 4-1/2 11.25 8.438 18.98 4.688 5.859 2.734

3 × 6 2-1/2 × 5-1/2 13.75 12.60 34.66 5.729 7.161 3.342

3 × 8 2-1/2 × 7-1/4 18.13 21.91 79.41 7.554 9.443 4.407

3 × 10 2-1/2 × 9-1/4 23.13 35.66 164.9 9.638 12.05 5.622

3 × 12 2-1/2 × 11-1/4 28.13 52.74 296.7 11.72 14.65 6.837

3 × 14 2-1/2 × 13-1/4 33.13 73.16 484.7 13.80 17.26 8.052

3 × 16 2-1/2 × 15-1/4 38.13 96.91 739.0 15.89 19.86 9.268

4 × 4 3-1/2 × 3-1/2 12.25 7.146 12.51 7.146 12.51 2.977

4 × 5 3-1/2 × 4-1/2 15.75 11.81 26.58 9.188 16.08 3.828

4 × 6 3-1/2 × 5-1/2 19.25 17.65 48.53 11.23 19.65 4.679

4 × 8 3-1/2 × 7-1/4 25.38 30.67 111.2 14.81 25.91 6.169

4 × 10 3-1/2 × 9-1/4 32.38 49.92 230.9 18.89 33.05 7.870

4 × 12 3-1/2 × 11-1/4 39.38 73.84 415.3 22.97 40.20 9.572

4 × 14 3-1/2 × 13-1/4 47.25 104.3 691.3 27.56 48.23 11.48

4 × 16 3-1/2 × 15-1/4 54.25 137.9 1051 31.65 55.38 13.19

5 × 5 4-1/2 × 4-1/2 20.25 15.19 34.17 15.19 34.17 4.922

6 × 6 5-1/2 × 5-1/2 30.25 27.73 76.26 27.73 76.26 7.352

6 × 8 5-1/2 × 7-1/2 41.25 51.56 193.4 37.81 104.0 10.03

6 × 10 5-1/2 × 9-1/2 52.25 82.73 393.0 47.90 131.7 12.70

6 × 12 5-1/2 × 11-1/2 63.25 121.2 697.1 57.98 159.4 15.37

6 × 14 5-1/2 × 13-1/2 74.25 167.1 1128 68.06 187.2 18.05

6 × 16 5-1/2 × 15-1/2 85.25 220.2 1707 78.15 214.9 20.72

6 × 18 5-1/2 × 17-1/2 96.25 280.7 2456 88.23 242.6 23.39

6 × 20 5-1/2 × 19-1/2 107.3 348.7 3400 98.36 270.5 26.08

6 × 22 5-1/2 × 21-1/2 118.3 423.9 4557 108.4 298.2 28.75

6 × 24 5-1/2 × 23-1/2 129.3 506.4 5950 118.5 325.9 31.43

8 × 8 7-1/2 × 7-1/2 56.25 70.31 263.7 70.31 263.7 13.67

8 × 10 7-1/2 × 9-1/2 71.25 112.8 535.9 89.06 334.0 17.32

8 × 12 7-1/2 × 11-1/2 86.25 165.3 950.5 107.8 404.3 20.96

8 × 14 7-1/2 × 13-1/2 101.3 227.9 1538 126.6 474.8 24.62

8 × 16 7-1/2 × 15-1/2 116.3 300.4 2328 145.4 545.2 28.27

8 × 18 7-1/2 × 17-1/2 131.3 383.0 3351 164.1 615.5 31.91

8 × 20 7-1/2 × 19-1/2 146.3 475.5 4636 182.9 685.8 35.56

8 × 22 7-1/2 × 21-1/2 161.3 578.0 6213 201.6 756.1 39.20

8 × 24 7-1/2 × 23-1/2 176.3 690.5 8113 220.4 826.4 42.85


WOOD PROPERTIES 403

Nominal

Size b × dStandard

Dressed Size


Area of

Section

A in2


in Pound per Linear


When Density

Equals 35 lb/ft3

Section

Modulus

SXX in3

Moment

of Inertia

IXX in4

Section

Modulus

SYY in3

Moment

of Inertia

IYY in4

10 × 10 9-1/2 × 9-1/2 90.25 142.9 678.8 142.9 678.8 21.94

10 × 12 9-1/2 × 11-1/2 109.3 209.5 1205 173.1 822.0 26.57

10 × 14 9-1/2 × 13-1/2 128.3 288.7 1949 203.1 964.9 31.18

10 × 16 9-1/2 × 15-1/2 147.3 380.5 2949 233.2 1108 35.80

10 × 18 9-1/2 × 17-1/2 166.3 485.0 4244 263.3 1251 40.42

10 × 20 9-1/2 × 19-1/2 185.3 602.2 5872 293.4 1394 45.04

10 × 22 9-1/2 × 21-1/2 204.3 732.1 7870 323.5 1537 49.66

10 × 24 9-1/2 × 23-1/2 223.3 874.6 10276 353.6 1679 54.27

12 × 12 11-1/2 × 11-1/2 132.3 253.6 1458 253.6 1458 32.16

12 × 14 11-1/2 × 13-1/2 155.3 349.4 2359 297.7 1712 37.75

12 × 16 11-1/2 × 15-1/2 178.3 460.6 3570 341.7 1965 43.34

12 × 18 11-1/2 × 17-1/2 201.3 587.1 5137 385.8 2218 48.93

12 × 20 11-1/2 × 19-1/2 224.3 729.0 7108 429.9 2472 54.52

12 × 22 11-1/2 × 21-1/2 247.3 886.2 9526 474.0 2725 60.11

12 × 24 11-1/2 × 23-1/2 270.3 1059 12439 518.1 2979 65.70

14 × 14 13-1/2 × 13-1/2 182.3 410.2 2769 410.2 2769 44.31

14 × 16 13-1/2 × 15-1/2 209.3 540.7 4190 470.9 3179 50.87

14 × 18 13-1/2 × 17-1/2 236.3 689.2 6031 531.7 3589 57.43

14 × 20 13-1/2 × 19-1/2 263.3 855.7 8343 592.4 3999 64.00

14 × 22 13-1/2 × 21-1/2 290.3 1040 11183 653.2 4409 70.56

14 × 24 13-1/2 × 23-1/2 317.3 1243 14602 713.9 4819 77.12

16 × 16 15- 1/2 × 15-1/2 240.3 620.8 4811 620.8 4811 58.41

16 × 18 15- 1/2 × 17-1/2 271.3 791.3 6924 700.9 5432 65.94

16 × 20 15- 1/2 × 19-1/2 302.3 982.5 9579 780.9 6052 73.48

16 × 22 15- 1/2 × 21-1/2 333.3 1194 12839 861.0 6673 81.01

16 × 24 15- 1/2 × 23-1/2 364.3 1427 16765 941.1 7294 88.55

18 × 18 17- 1/2 × 17-1/2 306.3 893.4 7817 893.4 7817 74.45

18 × 20 17- 1/2 × 19-1/2 341.3 1109 10815 995.5 8710 82.95

18 × 22 17- 1/2 × 21-1/2 376.3 1348 14495 1098 9603 91.46

18 × 24 17- 1/2 × 23-1/2 411.3 1611 18928 1200 10497 99.97

20 × 20 19- 1/2 × 19-1/2 380.3 1236 12051 1236 12051 92.43

20 × 22 19- 1/2 × 21-1/2 419.3 1502 16152 1363 13287 101.9

20 × 24 19- 1/2 × 23-1/2 458.3 1795 21091 1489 14522 111.4

22 × 22 21- 1/2 × 21-1/2 462.3 1657 17808 1657 17808 112.4

22 × 24 21- 1/2 × 23-1/2 505.3 1979 23254 1811 19465 122.8

24 × 24 23- 1/2 × 23-1/2 552.3 2163 25417 2163 25417 134.2


APPENDIX 7

FORMWORK CHARTS (WILLIAMS)

404


405


406


407


408


409


410


411

APPENDIX 8

FORM HARDWARE VALUES (WILLIAMS)

412

413

Shebolt Tie-Rod Forming System

SHEBOLT TIE-ROD FORMING SYSTEM 415

416 FORM HARDWARE VALUES (WILLIAMS)

EXTERNAL FASTENERS 417

External Fasteners

418 FORM HARDWARE VALUES (WILLIAMS)

EXTERNAL FASTENERS 419

420

Taper Tie Forming System

APPENDIX 9

ALUMINUM BEAMS (ALUMA)

422

ALUMINUM BEAMS (ALUMA) 423

Aluma Strongback Load ChartImperial*

Span (ft)Allowable

Deflection

L/360 (in)

1 Span (lbs/ft) 2 Span (lbs/ft) 3 Span (lbs/ft)

4.00 0.13 10006M 7603R 8640R

4.50 0.15 7906M 6758R 7680R

5.00 0.17 6404M 6082R 6912R

5.50 0.18 5101 D 5292M 6283R

6.00 0.20 3929 D 4447M 5559M

6.50 0.22 3090 D 3789M 4737M

7.00 0.23 2474 D 3267M 4084M

7.50 0.25 2012 D *2012 2846M 3558M

8.00 0.27 1657 D *1554 2502M 3127M *2932

8.50 0.28 1382 D *1219 2216M 2608 D *2301

9.00 0.30 1164 D *970 1977M 2197 D *1831

9.50 0.32 990 D *781 1774M 1868 D *1475

10.00 0.33 849 D *636 1601M *1533 1601 D *1201

10.50 0.35 733 D *524 1452M *1261 1383 D *988

11.00 0.37 638 D *435 1323M *1047 1203 D *820

11.50 0.38 558 D *364 1211M *877 1053 D *687

12.00 0.40 491 D *307 1112M *739 927 D *579

424 ALUMINUM BEAMS (ALUMA)

Aluma Beam Load ChartImperial*

Span (ft)Allowable

Deflection

L/360 (in)

1 Span (lbs/ft) 2 Span (lbs/ft) 3 Span (lbs/ft)

4.00 0.13 3151M** 2471R 2808R

4.50 0.15 2490M 2196R 2496R

5.00 0.17 2017M 1977R 2246R

5.50 0.18 1537 D 1797R 2042R

6.00 0.20 1184 D 1402M 1753M

6.50 0.22 931 D 1193M 1728R

7.00 0.23 745 D 1030M 1288M

7.50 0.25 606 D *606 896M 1144 D *1144

8.00 0.27 499 D *468 788M 942 D *883

8.50 0.28 416 D *367 676M 786 D *693

9.00 0.30 351 D *292 622M 662 D *551

9.50 0.32 298 D *235 558M 563 D *444

10.00 0.33 256 D *192 509M *462 482 D *362

This information is proprietary to Aluma Systems and is subject to change.It is intended to be used by technically skilled designers knowledgeable in thefield and is to be used with other data.

Trim Size: 6.125in x 9.25in Souder bindex.tex V1 - 09/12/2014 4:20pm Page 425

INDEX

A

A36 steel, 26, 105Access examples, 341ACI 347, 158Active soil coefficient, 82Active soil pressure, 82Active soil pressure diagrams, 90Alternate anchoring methods, 289Aluma beams, 205Aluma shoring, 245Aluminum beams, 205Ambient air temperature, 157American Concrete Institute (ACI),

155, 158American Institute of Steel Construction

(AISC), 25, 105Anchoring to concrete, 291Antlers bridge case study, 308Atterberg limit test, 76Axial, 47Axial force, 4

B

Basic building material weight, 345Beam design, 34

Beam software, 33, 120

Bearers, 62

Bearing capacity, 241

Bending moments, 13, 22

Bending stress, 19

Benicia-Martinez Bridge, 158

Bidwell Park, Chico, CA, 214

Boussinesq loads, 100

Box girder falsework design, 262

Box girder geometry, 262

Braced sheet piles, 138

Bridge falsework stringer design,

262

Bridge Project, 262

Buckling stress, 38, 66

C

Cantilevered formwork, 168

Cantilevered sheet piles, 136

Cantilevered shoring systems, 94

Cantilevered soldier beams, 115

Cap beam design, 351

Center of gravity, 4

Centroidal axis, 7

Centroids, 4

425


426 INDEX

Chrystal Springs Project, 174

CIDH anchor, 290

Coarse-grained soils, 76

Coefficient of friction, 274

Cold-rolled sheetpile, 131

Column forms, 165

Combined loads, 47

Combined stress, 43, 66

Compact beam, 25

Composite formwork design, 202

Compression, 4, 47

Concentrated loads, 4, 47

Concrete chemical coefficient,

163

Concrete deadmen, 269, 277,

Concrete finish costs, 222

Concrete Formula #1 and #2, 165

Concrete mix design, 161

Concrete placement costs, 222

Concrete placement height, 162

Concrete placement methods, 161

Concrete placement rate, 160

Concrete pressure, 51

Concrete pressure diagrams, 165

Concrete pressure underwater, 173

Concrete Specifications, 178

Concrete submittals, 179

Concrete temperature, 157

Concrete unit weight coefficient,

162

Corewall formwork, 251

Coulomb, 91, 124

Counterweights, 70

Crane trestle, 322

Cross-sectional area, 7

Curved failure planes, 91

D

Deflection, 27

Dimensional lumber, 26

Distributed loads, 22

Double steel channels, 358

Drilled pile anchor, 290

Drilled-in concrete anchors, 291

Drop beam falsework design, 244

Dry unit weight, 79

E

End connections, 39End connections, 46Engineer falsework reviews, 234Estimating concrete formwork, 219Estimating with formwork ratios, 226Excavation sequencing, 131Execution of work submittals, 179Existing structure stringer design, 350Existing structure weight, 349, 356Existing water treatment plant, 347External walers, 143

F

Factor of safety, 24, 62Falsework accidents, 230Falsework bracing, 265Falsework design criteria, 235Falsework lateral force, 266Falsework load paths, 236Falsework loads, 175Falsework post design, 240Falsework removal, 176Falsework review process, 233Falsework risks, 229Fine-grained soils, 76Flange buckling, 304, 318Flat slab falsework, 236Flowing water pressure, 48Fly tables, 249Folsom Dam, 163Form bracing, 269Form bracing design, 270Form material submittals, 179Form ratio, 29Formwork adjustment factors, 184Formwork allowable stresses, 184Formwork charts, 199Formwork charts, 242Formwork costs, 179Formwork design, 180, 185Formwork drawings, 213Formwork estimating, 219Formwork geometry, 213Formwork labor costs, 221Formwork load path, 183


INDEX 427

Formwork ratios, 219

Formwork removal, 176

Formwork stud design, 188

Formwork tie capacity, 198

Formwork waler design, 188

Free-body diagrams (FBD), 16

Friction force, 275

Full liquid pressure, 167

G

Galena Creek Bridge, 341

Golden State Bridge, 336

Grades of steel, 24

Guying pier forms, 278

H

H pile strut, 148

Hanging scaffold, 69

Hard soils, 136

HDO plywood, 207

Heavy duty, 62

Highway 149 accident, 231

Historical formwork cost data, 220

Horizontal shear stress, 20

Hot-rolled sheet pile, 131

Hurd, M. K., 155

Hydraulic crane trestle design, 327

Hydraulic hammers, 134

Hydrostatic pressure, 53, 157, 164

I

Impact (Diesel) hammers, 134

Impact factor (IF), 307, 330

Imperial units, 1

Internal bracing, 292

Internal walers, 142

IWRC, 71

L

Lagging, 105, 108

Lateral pressure adjustments, 162

Law of superposition, 23

Laws of equilibrium, 14

Light duty, 62

Loading submittals, 179

Loads on sloped surfaces, 51

Log of test∖borings, 77Longitudinal bracing, 295

M

Medium duty, 62

Modulus of elasticity, 11, 22

Mohr, 124

Moment of inertia, 7

Multibraced shoring, 97

N

National Design Specifications (NDS),

106

Normal force, 275

Normal stress, 18

O

Occupational Safety and Health

Administration (OSHA), 59

One-sided forms, 172

Ord Ferry Bridge, 336

OSHA Trench Classifications, 100

Owner falsework reviews, 234

P

Pankow Builders, LTD., 248

Pankow,Charles, 248

Passive soil coefficient, 83

Passive soil pressure, 83

Pile caps, 362

Pile design, trestle, 335

Pile driving equipment, 133

Pile load, 333

Pipe strut, 148

Pipe supports, 355

Pipe weight, 346

Planking, 62

Plasticity index (PI), 76

Pleasant Grove WWTP, 211

Plywood, 26

Plywood design, 186

Post design, 265


428 INDEX

Post-tensioning, 177

Precast concrete trestle decking, 306

Premanufactured scaffold frames, 59

R

Radius of gyration, 11

Rankine theory, 83

Reactions, 4

Rebar bracing, 268

Rebar bracing case study, 297

Rebar guying, highway, 279

Reinforcing steel, 156

Reinforcing steel costs, 224

Resultant forces, 4, 13

River forces, 334

Rule of eighths, 286

Runners, 62

S

Safe Working Load (SWL), 123

Sample soil strength, 124

Scaffold design, 63

Scaffold loads, 61

Scaffold posts, 63

Section modulus, 9

Securing scaffold, 69

Shear and moment diagrams

(SMD), 28

Shear strength, 124

Shear stress, 19

Sheet piling, 130, 131

Shoring costs, 152

Shoring frames, 246

Short-term loading, 106

Simply supported, 45

Slip planes, 86

Small tools, services and supplies

(STS), 223

Soil anchor design, 125

Soil anchors, 105

Soil arching, 107

Soil mechanics, 81

Soil nail design, 125

Soil nails, 105, 121, 123

Soil pressure, 55

Soil Pressure, 83

Soil pressure formulas, 93

Soil properties, 75

Soil shear strength, 79

Soil testing, 81

Soil unit weight, 80

Soldier beams, 104, 112

Specific gravity, 79

Standard Penetration Test, 77

Star pile, 340

Statics, 1

Steam hammers, 134

Steel beam design, 35

Steel channel form design, 208

Steel column design, 41

Steel plate lagging, 111

Stress, 18

Strut design, 146

Struts, 97, 132, 145

Sub-cap beam design, 351

Support of excavation methods, 128

Supported shoring systems, 94

Supported soldier beam system, 117

Supporting existing structures, 57

Surcharge loads, 98

Suspended scaffold, 70

T

Templates, 135

Temporary pipe supports, 354

Temporary supports, 344

Tension, 4, 47

Terzaghi, 91

Tiebacks, 96

Tiebacks, 105, 121

Timber trestle decking, 306, 324

Trackingproject costs, 227

Trans Bay Terminal project (SF), 342

Trench sloping, 101

Tresles, 300

Trestle cap beams, 301, 330

Trestle decking, 306

Trestle design, 312

Trestle environmental considerations,

308

Trestle foundation, 301


INDEX 429

Trestle lateral bracing, 303

Trestle pile, 331

Trestle stringers, 303, 325

Tributary width, 22

Tube and coupler scaffold, 59

Types of concrete structures, 225

U

Ultimate stress, 25

Unified Soils Classification

System, 76

Uniform load, 5

Uniform loads, 47

Unit weights, 155

Units of measure, 1

V

Velocity of water, 334

Vibratory hammers, 133

Void ratio, 78

W

Walers, 96, 132, 140

Water pressure, 48

Water table affects, 98

Web yielding, 150

Wind force, 270, 280

Wind forces, 53

Wire rope capacities, 283, 358

Wire rope clips, 284

Wire rope termination lengths, 287

Wood beam design, 36

Wood form design, 189

Wood lagging design, 108

Wood Species, 106

X

XIPS, 285

Y

Yield stress, 25

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